作者：w703710691d

Ubuntu 18.04 下 MySQL 8.0 配置远程访问

先执行

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sudo netstat -nalpt | grep 0.0.0.0:3306

查看是否3306端口已被监听

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tcp        0      0 0.0.0.0:3306            0.0.0.0:*               LISTEN      24283/mysqld

如果不存在先执行以下步骤：

打开文件 /etc/mysql/mysql.conf.d/mysqld.cnf

在最后加入一行内容 

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bind-address    = 0.0.0.0

保存后重启mysql服务器

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sudo service mysql restart

然后再查看是否3306端口已被监听

 

Codeforces 847J Students Initiation 二分答案 最大流

Codeforce 847J Students Initiation

题意

给一个包含$n$($n \leq 5000$)个点的无向图，边的条数$m \leq 5000$，没有重边自环。现在给每条边一个方向，问每个点的出度的最大值，最小为多少。输出这个最小值，并输出方案。

思路&建图

  1. 看到最大值最小，最小值最大，首先想到炸天哥猜想：二分！！
  2. 二分这个最大值$mid$，检验这个猜测值是否满足要求
  3. 从源点相这$n$个点引一条边，容量为$mid$。
  4. 对于第$i$条边来说，其编号为$n+i$，从其两个断点分别引一条边到这个点，容量为$1$。再从这个点到汇点连一条边，容量为$1$。
  5. 跑最大流，如果最大流和边的数目相同，即这个猜测值可行。此时，对于每一条边对应的点来说，这条边的方向就应该是从[流量来的那个点]到[没有流量流过来的那个点]。

代码

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 5;
const int maxm = 1e6;
struct EDGE
{
    int to, next, flow;
    EDGE(int to = 0, int next = 0, int flow = 0) : to(to), next(next), flow(flow) {}
} edge[maxm];
int head[maxn], edgecnt, maxNodeNum;
void init()
{
    memset(head, -1, sizeof(head));
    edgecnt = 0;
    maxNodeNum = 0;
}
void AddEdge(int s, int t, int c)
{
    edge[edgecnt] = EDGE(t, head[s], c);
    head[s] = edgecnt++;
    edge[edgecnt] = EDGE(s, head[t], 0);
    head[t] = edgecnt++;
    maxNodeNum = max(maxNodeNum, max(s, t));
}
int layer[maxn];
int p[maxn];
bool GetLayer(int st, int en)
{
    memset(layer, 0, sizeof(int) * (maxNodeNum + 1));
    layer[st] = 1;
    queue<int>q;
    q.push(st);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u]; ~i; i = edge[i].next)
        {
            if(edge[i].flow == 0)
                continue;
            int v = edge[i].to;
            if(layer[v])
                continue;
            layer[v] = layer[u] + 1;
            q.push(v);
        }
    }
    return layer[en];
}
int dfsFlow(int u, int en, int f)
{
    int ret = 0;
    if(u == en)
        return f;
    if(layer[u] >= layer[en])
        return 0;
    for(; ~p[u]; p[u] = edge[p[u]].next)
    {
        if(edge[p[u]].flow == 0)
            continue;
        int v = edge[p[u]].to;
        if(layer[v] != layer[u] + 1)
            continue;
        int temp = dfsFlow(v, en, min(f, edge[p[u]].flow));
        ret += temp;
        f -= temp;
        edge[p[u]].flow -= temp;
        edge[p[u] ^ 1].flow += temp;
        if(f == 0)
            break;
    }
    return ret;
}
int dinic(int st, int en)
{
    int ans = 0;
    while(GetLayer(st, en))
    {
        memcpy(p, head, sizeof(int) * (maxNodeNum + 1));
        ans += dfsFlow(st, en, INT_MAX);
    }
    return ans;
}
pair<int, int> E[maxn];
int n, m;
bool check(int mid)
{
    init();
    for(int i = 1; i <= n; i++)
        AddEdge(0, i, mid);
    for(int i = 1; i <= m; i++)
    {
        AddEdge(E[i].first, n + i, 1);
        AddEdge(E[i].second, n + i, 1);
    }
    int en = n + m + 1;
    for(int i = n + 1; i < en; i++)
        AddEdge(i, en, 1);
    int ret = dinic(0, en);
    return ret == m;
}
int main()
{
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= m; i++)
        scanf("%d %d", &E[i].first, &E[i].second);
    int L = 0, R = m;
    while(L < R)
    {
        int mid = L + R >> 1;
        if(check(mid))
            R = mid;
        else
            L = mid + 1;
    }
    check(L);
    printf("%d\n", L);
    for(int i = 1; i <= n; i++)
    {
        for(int j = head[i]; ~j; j = edge[j].next)
        {
            if(j & 1)
                continue;
            if(edge[j].flow)
                continue;
            int eid = edge[j].to - n;
            int v;
            if(E[eid].first == i)
                v = E[eid].second;
            else
                v = E[eid].first;
            printf("%d %d\n", i, v);
        }
    }
    return 0;
}

 

 

HDU 6214 Smallest Minimum Cut 最小割的割边的数目最少

Smallest Minimum Cut

题意

给一个有向的网络图，每条边有一定的容量$w_i$，现在以$1$点为源点，$n$点为汇点。求一组最小割的割边，使边的数目最少，输出最少数目。

建图&思路

直接按照原网络建图，只是边的容量改变为$10000 * w_i +1$，其中边的条数$m<10000$。这样跑最大流（最小割）后，假设最大流为$flow$，那么$flow / 10000$就是原始网络的最大流（最小割），$flow \mod 10000$就是最小割的最少边的条数。

代码

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 300;
const int maxm = 4000;
struct EDGE
{
    int to, next, flow;
    EDGE(int to = 0, int next = 0, int flow = 0): to(to), next(next), flow(flow) {}
} edge[maxm];
int head[maxn], edgecnt, maxNodeNum;
void init()
{
    memset(head, -1, sizeof(head));
    edgecnt = 0;
    maxNodeNum = 0;
}
void AddEdge(int s, int t, int c)
{
    edge[edgecnt] = EDGE(t, head[s], c);
    head[s] = edgecnt++;
    edge[edgecnt] = EDGE(s, head[t], 0);
    head[t] = edgecnt++;
    maxNodeNum = max(maxNodeNum, max(s, t));
}
int layer[maxn];
int p[maxn];
bool GetLayer(int st, int en)
{
    memset(layer, 0, sizeof(int) * (maxNodeNum + 1));
    layer[st] = 1;
    queue<int>q;
    q.push(st);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u]; ~i; i = edge[i].next)
        {
            if(edge[i].flow == 0)
                continue;
            int v = edge[i].to;
            if(layer[v])
                continue;
            layer[v] = layer[u] + 1;
            q.push(v);
        }
    }
    return layer[en];
}
int dfsFlow(int u, int en, int f)
{
    int ret = 0;
    if(u == en)
        return f;
    if(layer[u] >= layer[en])
        return 0;
    for(; ~p[u]; p[u] = edge[p[u]].next)
    {
        if(edge[p[u]].flow == 0)
            continue;
        int v = edge[p[u]].to;
        if(layer[v] != layer[u] + 1)
            continue;
        int temp = dfsFlow(v, en, min(f, edge[p[u]].flow));
        ret += temp;
        f -= temp;
        edge[p[u]].flow -= temp;
        edge[p[u] ^ 1].flow += temp;
        if(f == 0)
            break;
    }
    return ret;
}
int dinic(int st, int en)
{
    int ans = 0;
    while(GetLayer(st, en))
    {
        memcpy(p, head, sizeof(int) * (maxNodeNum + 1));
        ans += dfsFlow(st, en, INT_MAX);
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        init();
        int n, m;
        scanf("%d %d", &n, &m);
        int s, t;
        scanf("%d %d", &s, &t);
        while(m--)
        {
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
            AddEdge(u, v, w * 10000 + 1);
        }
        int ret = dinic(s, t) % 10000;
        printf("%d\n", ret);
    }
    return 0;
}

 

计蒜客 Our Journey of Dalian Ends 费用流找最短路

Our Journey of Dalian Ends

题意

给一个无向图，有边权，每个点用地名字符串表示。现在要从大连出发到达西安，并且必须经过上海，且给个点只能经过一次，问最短路径。

建图&思路

  1. 可以认为是找两条路径，起点都是上海，分别到达西安和大连，且经过的点不能重复，求路径和的最小长度。
  2. 将每个点拆点，拆分为入点和出点，入点向出点连一条边，容量为$1$，费用为$0$，代表这个点只能经过一次。
  3. 对于原图上的点，从一个城市的出点连一条边，到另一个城市所代表的点的入点。容量为$1$，费用为这条边的边权。反过来也需要连一条边，代表双向可行。
  4. 从源点向上海的出点连一条边，容量为$2$，费用为$0$，代表需要找到两条路径。
  5. 从西安和大连的出点，分别引一条边到汇点，容量为$1$，费用为$0$，代表只能是以这两个点为终点。
  6. 跑费用流，如果流量为2，那么费用就是两条路径的最短长度。
  7. 当然，如果输入数据中没有这三个城市，直接不行啊！

代码

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#include<bits/stdc++.h>
const int maxn = 100005;
#define inf 0x3FFFFFFF
using namespace std;
struct node
{
int st;
int en;
int flow, cost;
int next;
} E[1001000];
int num;
int p[maxn];
void init()
{
memset(p, -1, sizeof p);
num = 0;
}
void add(int st, int en, int flow, int cost)
{
E[num].st = st;
E[num].en = en;
E[num].flow = flow;
E[num].cost = cost;
E[num].next = p[st];
p[st] = num++;
E[num].st = en;
E[num].en = st;
E[num].flow = 0;
E[num].cost = -cost;
E[num].next = p[en];
p[en] = num++;
}
int n;
 
int pre[maxn];
int dis[maxn];
bool fg[maxn];
 
 
bool spfa(int st, int en)
{
for (int i = 0; i <= en; i++)
fg[i] = 0, dis[i] = inf, pre[i] = -1;
queue<int>q;
q.push(st);
fg[st] = 1;
dis[st] = 0;
while (!q.empty())
{
int u = q.front();
q.pop();
fg[u] = 0;
for (int i = p[u]; i + 1; i = E[i].next)
{
int v = E[i].en;
if (E[i].flow && dis[v] > dis[u] + E[i].cost)
{
dis[v] = dis[u] + E[i].cost;
pre[v] = i;
if (!fg[v])
{
fg[v] = 1;
q.push(v);
}
}
}
}
if (dis[en] < inf)
return 1;
return 0;
}
pair<int, int> solve(int st, int en)
{
int ans = 0;
int f = 0;
while (spfa(st, en))
{
int d = inf;
for (int i = pre[en]; i + 1; i = pre[E[i].st])
d = min(d, E[i].flow);
for (int i = pre[en]; i + 1; i = pre[E[i].st])
{
E[i].flow -= d;
E[i ^ 1].flow += d;
ans += d * E[i].cost;
}
f += d;
}
return make_pair(ans, f);
}
char from[maxn], to[maxn];
vector<pair<int, int> > vec[maxn];
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
for (int i = 0; i < maxn; i++)
vec[i].clear();
int cnt = 0;
init();
        unordered_map<string, int> mp;
int m;
scanf("%d",&m);
while (m--)
{
int len;
scanf("%s %s %d", from, to, &len);
if (mp[from] == 0)
mp[from] = ++cnt;
if (mp[to] == 0)
mp[to] = ++cnt;
int u = mp[from];
int v = mp[to];
vec[u].emplace_back(v, len);
vec[v].emplace_back(u, len);
}
if(!mp.count("Dalian") || !mp.count("Shanghai") || !mp.count("Xian"))
        {
            puts("-1");
            continue;
        }
for (int u = 1; u <= cnt; u++)
{
    if(u != mp["Shanghai"])
                add(u, cnt + u, 1, 0);
for (auto &p : vec[u])
{
int v = p.first;
int len = p.second;
add(cnt + u, v, 1, len);
}
}
add(0, mp["Shanghai"] + cnt, 2, 0);
add(mp["Dalian"] + cnt, cnt + cnt + 1, 1, 0);
add(mp["Xian"] + cnt, cnt + cnt + 1, 1, 0);
pair<int, int> ret = solve(0, cnt + cnt + 1);
if (ret.second != 2)
{
puts("-1");
}
else
{
printf("%d\n", ret.first);
}
}
return 0;
}

 

 

Codeforces 546E Soldier and Traveling 最大流找可行方案

Codeforces 546E

题意

一个国家有$n$个城市，由$m$条双向边连接。现在，每个城市有$a_i$个士兵。现在每个士兵可以沿着一条边，走到相邻的城市，并且只能走一次，或者不走，留在原来的城市。问能否达到每个城市分别有$b_i$的士兵。切每条道路有一个上限，通过的士兵的个数不能超过这个上限。

  • 如果可以，输出"YES"。并输出方案，用矩阵$A_{ij}$表示从$i$城市走到$j$城市的士兵的个数。
  • 如果不可以，输出"NO"。

建图&思路

  1. 左边$n$个点，从源点往分别这$n$个点引一条边，容量为$a_i$。
  2. 右边$n$个点，从这个$n$个点分别往汇点引一条边，容量为$b_i$。
  3. 对于每一个城市$i$来说，从左边的第$i$个点向右边的第$i$个点引一条边，容量为$INF$，表示允许驻留。
  4. 对于每一条路来说，它的两个断点分别为$st,en$，那么需要从左边的第$st$个点引一条边到右边的第$en$个点，容量为这条路的上限。同时，从左边的第$en$个点引一条边到右边的第$st$个点，容量为这条路的上限。表示这两个城市之间的士兵可以相互移动，数量的上限不超过路的上限。
  5. 跑最大流，得到最大流量为$flow$，如果$flow$和士兵的总数相同，那么就有可行方案。否则没有。
  6. 对于每一个左边的点，枚举每一条向右边连的边，那么从这个点对应的城市到所连的点的城市所移动的士兵的个数，就是这条边所流过的流量，即这条边反向边的残余流量。

代码

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#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
using namespace std;
#define INFF 0x3fffffff
#define MM 100010
#define NN 5100
struct node
{
    int st, en, flow;
    int next;
} E[MM];
int num;
int p[NN];
void init()
{
    memset(p, -1, sizeof(p));
    num = 0;
}
void add(int st, int en, int flow)
{
    E[num].st = st;
    E[num].en = en;
    E[num].flow = flow;
    E[num].next = p[st];
    p[st] = num++;
    E[num].st = en;
    E[num].en = st;
    E[num].flow = 0;
    E[num].next = p[en];
    p[en] = num++;
}
int T;
int vdis[NN];//到汇点的距离为i（即是同一层）个数为多少
int dis[NN];//i点到汇点距离
int Find(int u, int st, int en, int f)
{
    //f为起点来的流量
    if (u == en)
        return f;//如果点相同返回起点来的流量
    int left = f; //剩余流量
    int tmp = T; //T是到汇点无限远距离
    for (int i = p[u]; i + 1; i = E[i].next)
    {
        int v = E[i].en;
        int flow = E[i].flow;
        if (flow) //如果剩余容量大于0
        {
            if (dis[v] + 1 == dis[u])
            {
                //u->v  倒着推
                int d = min(left, flow); //这条边能过去的最大流量
                d = Find(v, st, en, d); //当前最大流量传向上一条边
                left -= d; //剩余多少
                E[i].flow -= d; //这条边的剩余容量
                E[i ^ 1].flow += d; //反向边的剩余容量
                if (dis[st] >= T + 1) //如果起点不能到汇点，即是距离大于等于无穷
                    return f - left; //返回流了多少流量
                if (!left) //如果剩余容量为0
                    break;
            }
            if (dis[v] < tmp) //下一条边的到汇点距离
                tmp = dis[v];
        }
    }
    if (left == f)
    {
        vdis[dis[u]]--;
        if (vdis[dis[u]] == 0)
            dis[st] = T + 1; //这一层没有点了
        dis[u] = tmp + 1;
        vdis[dis[u]]++;
    }
    return f - left;
}
int isap(int st, int en, int flow)
{
    memset(vdis, 0, sizeof(vdis));
    memset(dis, 0, sizeof(dis));
    vdis[0] = T + 1;
    int ans = 0;
    while (dis[st] <= T)
    {
        int tmp = Find(st, st, en, flow);
        ans += tmp;
        flow -= tmp;
        if (!flow)
            return ans;
    }
    return ans;
}
int A[205][205];
int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    init();
    T = 2 * n + 5;
    int sum = 0;
    for (int i = 1; i <= n; i++)
    {
        int x;
        scanf("%d", &x);
        sum += x;
        add(0, i, x);
    }
    int sum2 = 0;
    for (int i = 1; i <= n; i++)
    {
        int x;
        scanf("%d", &x);
        sum2 += x;
        add(i + n, n + n + 1, x);
        add(i, n + i, INFF);
    }
    if (sum2 != sum)
    {
        puts("NO");
        return 0;
    }
    for (int i = 0; i < m; i++)
    {
        int st, en;
        scanf("%d %d", &st, &en);
        add(st, en + n, INFF);
        add(en, st + n, INFF);
    }
    int t = isap(0, 1 + n + n, INFF);
    if (t != sum) printf("NO\n");
    else
    {
        puts("YES");
        memset(A, 0, sizeof(A));
        for (int i = 1; i <= n; i++)
        {
            for (int j = p[i]; ~j; j = E[j].next)
                A[i][E[j].en - n] = E[j ^ 1].flow;
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++) printf("%d ", A[i][j]);
            printf("\n");
        }
    }
    return 0;
}

注释

这个代码是2015年写的isap模板，所以不怎么好看。

Codeforces 863F 费用流求最小花费方案

Codeforces 863F

题意

  • 构造一个长度为$n$的序列$a_i$（$1 \leq n \leq 50, 1 \leq a_i \leq n$），需要满足$q$个条件（$0 \leq q \leq 100$）
  • 对于每一个条件，有一下两种可能
    1 $l_i,r_i,v_i$，对于所有满足$l_i \leq x \leq r_i$的下标$x$，有$a_x \geq v_i$
    2 $l_i,r_i,v_i$，对于所有满足$l_i \leq x \leq r_i$的下标$x$，有$a_x \leq v_i$
  • 构造出该序列的花费$cost = \sum_{i=1}^{n}cnt(i)^2$，其中$cnt(i)$为数$i$在构造的序列中出现的次数，求最小的花费。

建图

  • 先确定每个位置能填的数的范围，如果范围非法，可直输出$-1$并结束
  • 左边$n$个点，分别代表$1~n$这$n$个位置上的数，从源点引一条边，到左边的每一个点，容量为$1$，费用为$0$。
  • 右边也有$n$个点，分别代表填写的数。对于每个点，向汇点引$n$条边，第$j$条边的容量为$1$，费用为$j^2-(j-1)^2$。
  • 对于左边的每个点，向范围内的数对应的右边的数连边，容量为$1$，费用为$0$。
  • 由于费用流会优先使用费用较小的边，所以对于一个单独的数对应的右边的点来说，他向汇点的边的，只会使用前几条，这几条边的费用加起来正好是通过这个点的流量的平方，满足题目要求。
  • 跑费用流，即可解决问题。

代码

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 200 + 5;
const int maxm = 1e5 + 5;
struct EDGE
{
    int to, next, flow, cost;
    EDGE() {}
    EDGE(int to, int next, int flow, int cost) : to(to), next(next), flow(flow), cost(cost) {}
} edge[maxm];
int head[maxn], edgecnt, maxNodeNum;
void init()
{
    memset(head, -1, sizeof(head));
    edgecnt = 0;
    maxNodeNum = 0;
}
void AddEdge(int st, int en, int flow, int cost)
{
    edge[edgecnt] = EDGE(en, head[st], flow, cost);
    head[st] = edgecnt++;
    edge[edgecnt] = EDGE(st, head[en], 0, -cost);
    head[en] = edgecnt++;
    maxNodeNum = max(maxNodeNum, max(st, en));
}
int pre[maxn], dis[maxn];
bool SPFA(int st, int en)
{
    static bool vis[maxn];
    memset(dis, 0x7F, sizeof(int) * (maxNodeNum + 1));
    dis[st] = 0;
    queue<int> q;
    q.push(st);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; ~i; i = edge[i].next)
        {
            int v = edge[i].to;
            if(edge[i].flow && dis[v] > dis[u] + edge[i].cost)
            {
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
            }
        }
    }
    return dis[en] != 0x7F7F7F7F;
}
pair<int, int> solve(int st, int en, int maxFlow = INT_MAX)
{
    int cost = 0, flow = 0;
    while(SPFA(st, en))
    {
        int d = maxFlow - flow;
        for(int u = en; u != st; u = edge[pre[u] ^ 1].to)
            d = min(d, edge[pre[u]].flow);
        flow += d;
        for(int u = en; u != st; u = edge[pre[u] ^ 1].to)
        {
            edge[pre[u]].flow -= d;
            edge[pre[u] ^ 1].flow += d;
            cost += edge[pre[u]].cost * d;
        }
        if(flow == maxFlow)
            break;
    }
    return make_pair(cost, flow);
}
bool area[55][55];
int main()
{
    init();
    int n, q;
    scanf("%d %d", &n, &q);
    for(int i = 1; i <= n; i++)
    {
        AddEdge(0, i, 1, 0);
        for(int j = 1; j <= n; j++)
            AddEdge(n + i, n + n + 1, 1, j * j - (j - 1) * (j - 1));
    }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            area[i][j] = 1;
    while(q--)
    {
        int op, l, r, v;
        scanf("%d %d %d %d", &op, &l, &r, &v);
        if(op == 1)
            for(int j = l; j <= r; j++)
                for(int k = 1; k < v; k++)
                    area[j][k] = 0;
        else
            for(int j = l; j <= r; j++)
                for(int k = v + 1; k <= n; k++)
                    area[j][k] = 0;
    }
    for(int i = 1; i <= n; i++)
    {
        bool ok = 0;
        for(int j = 1; j <= n; j++)
            if(area[i][j])
            {
                AddEdge(i, n + j, 1, 0);
                ok = 1;
            }
        if(!ok)
        {
            puts("-1");
            return 0;
        }
    }
    int ans = solve(0, n + n + 1).first;
    printf("%d\n", ans);
    return 0;
}

 

CodeForces 818G 费用流 优化建图

CodeForces 818G

题意

给定一个长度为$n$的序列$a_i$，$4 \leq n \leq 3000,1 \leq a_i \leq 10^5$，从中选出$4$个不想交的子序列，满足每个子序列中相邻的两个项，要么是差的绝对值为$1$，要么他们模$7$的结果相同。问这四个序列的总长度最长为多少。

思路

  • 对于每个项来说，将其看成一个点，将其拆点为为$i$和$i+n$，从$i$向$i+n$引一条边，容量为$1$，费用为$-1$，表示该点项只能被选一次，且选了之后收益$+1$。这一对点的流量方向，一定为从$i$进，并且从$i+n$出。
  • 对于每个项来说，都有可能作为子序列的起点或者终点。所以，从源点引一条边到$i$点，容量为$1$，费用为$0$；再从$i+n$点引一条边到汇点，容量为$1$，费用为$0$。
  • 对于每一个可能相邻的项$a_i,a_j$且$i \lt j$来说，需要建一条边从$i+n$到$j$，容量为$1$，费用为$-1$。但是这样建边之后，会发现边很多，根本跑不过。其实会发现，对于每一类相邻来说，只需要往后找到四个是同类的并建边就好了。
  • 从源点压入$4$点流量，跑费用流即可，答案为最小费用的相反数。

代码

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 6e3 + 5;
const int maxm = 5e5 + 5;
struct EDGE
{
    int to, next, flow, cost;
    EDGE() {}
    EDGE(int to, int next, int flow, int cost) : to(to), next(next), flow(flow), cost(cost) {}
} edge[maxm];
int head[maxn], edgecnt, maxNodeNum;
void init()
{
    memset(head, -1, sizeof(head));
    edgecnt = 0;
    maxNodeNum = 0;
}
void AddEdge(int st, int en, int flow, int cost)
{
    edge[edgecnt] = EDGE(en, head[st], flow, cost);
    head[st] = edgecnt++;
    edge[edgecnt] = EDGE(st, head[en], 0, -cost);
    head[en] = edgecnt++;
    maxNodeNum = max(maxNodeNum, max(st, en));
}
int pre[maxn], dis[maxn];
bool SPFA(int st, int en)
{
    static bool vis[maxn];
    memset(dis, 0x7F, sizeof(int) * (maxNodeNum + 1));
    dis[st] = 0;
    queue<int> q;
    q.push(st);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; ~i; i = edge[i].next)
        {
            int v = edge[i].to;
            if(edge[i].flow && dis[v] > dis[u] + edge[i].cost)
            {
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
            }
        }
    }
    return dis[en] != 0x7F7F7F7F;
}
pair<int, int> solve(int st, int en, int maxFlow = INT_MAX)
{
    int cost = 0, flow = 0;
    while(SPFA(st, en))
    {
        int d = maxFlow - flow;
        for(int u = en; u != st; u = edge[pre[u] ^ 1].to)
            d = min(d, edge[pre[u]].flow);
        flow += d;
        for(int u = en; u != st; u = edge[pre[u] ^ 1].to)
        {
            edge[pre[u]].flow -= d;
            edge[pre[u] ^ 1].flow += d;
            cost += edge[pre[u]].cost * d;
        }
        if(flow == maxFlow)
            break;
    }
    return make_pair(cost, flow);
}
vector<int> lastPos[maxm];
vector<int> lastModPos[7];
int a[maxn];
int main()
{
    init();
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    for(int i = n; i >= 1; i--)
    {
        AddEdge(0, i, 1, 0);
        AddEdge(i + n, n + n + 1, 1, 0);
        AddEdge(i, i + n, 1, -1);
        for(int j = 0; j < min(4, (int)lastPos[a[i] + 1].size()); j++)
        {
            int sz = lastPos[a[i] + 1].size();
            AddEdge(i + n, lastPos[a[i] + 1][sz - j - 1], 1, 0);
        }
        for(int j = 0; j < min(4, (int)lastPos[a[i] - 1].size()); j++)
        {
            int sz = lastPos[a[i] - 1].size();
            AddEdge(i + n, lastPos[a[i] - 1][sz - j - 1], 1, 0);
        }
        for(int j = 0; j < min(4, (int)lastModPos[a[i] % 7].size()); j++)
        {
            int sz = lastModPos[a[i] % 7].size();
            AddEdge(i + n, lastModPos[a[i] % 7][sz - j - 1], 1, 0);
        }
        lastPos[a[i]].push_back(i);
        lastModPos[a[i] % 7].push_back(i);
    }
    int ret = -solve(0, n + n + 1, 4).first;
    printf("%d\n", ret);
    return 0;
}

2014-2015 ACM-ICPC Southwestern Europe Regional Contest (SWERC 14)

https://odzkskevi.qnssl.com/727b6827e404fd453ff4858e2619c49c?v=1504751451

A - GREAT+SWERC=PORTO

全排列枚举。先将字母按照出现的顺序，重新编码为以$0$为开始值的数字，方便后续使用。然后构造一个排列数组$p$，长度为$10$，初始值$p[i] = i$。然后枚举$p$的全排列，$p[i]$的意义就是将数字（实质上是被重新编码后的字母）$i$以值$p[i]$来代替。假设有$cnt$个不同的字母，全排列我们枚举的却是$10$个数，每次只取前$cnt$个进行代替，那么最终的答案会重复计算$(10-cnt)!$次。进行暴力枚举并统计答案后，除以$(10-cnt)!$即可。

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#include<bits/stdc++.h>
using namespace std;
char str[12][12];
int length[12];
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        for(int i = 1; i <= n; i++)
            scanf("%s", str[i]);
        int mp[127];
        memset(mp, -1, sizeof(mp));
        int cnt = 0;
        for(int i = 1; i <= n; i++)
        {
            length[i] = strlen(str[i]);
            for(int j = 0; j < length[i]; j++)
            {
                if(mp[str[i][j]] != -1)
                    continue;
                mp[str[i][j]] = cnt++;
            }
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 0; j < length[i]; j++)
                str[i][j] = mp[str[i][j]];
        }//对字母进行重新编码
        int p[12];
        for(int i = 0; i < 10; i++)
            p[i] = i;//排列数组
        int ans = 0;
        do
        {
            bool ok = 1;
            for(int i = 1; i <= n; i++)
            {
                if(p[str[i][0]] == 0)//不能将第一位替换为0
                {
                    ok = 0;
                    break;
                }
            }
            if(!ok)
                continue;
            int val[12];
            memset(val, 0, sizeof(val));
            for(int i = 1; i <= n; i++)
            {
                for(int j = 0; j < length[i]; j++)
                    val[i] = val[i] * 10 + p[str[i][j]];//计算替换后，每个单词的值
            }
            int sum = 0;
            for(int i = 1; i < n; i++)
                sum += val[i];
            if(sum == val[n])
                ans++;
        }
        while(next_permutation(p, p + 10));//枚举下一个排列
        int t = 1;
        for(int i = 10 - cnt; i >= 2; i--)
            t *= i;//(10-cnt)的阶乘
        printf("%d\n", ans / t);
    }
    return 0;
}

B - Flowery Trails

从起点和终点进行两次最短路。开一个$dis[2][i]$数组，分别表示从$0$点到$i$点的最短路、从$n-1$点到$i$点的最短路。要判断一条边是否在最短路上，可以枚举每一条$单向边$，假设这条边的两个端点分别为$u$和$v$，如果满足$dis[0][u] + len + dis[1][v] = dis[0][n-1]$，那么从$u->v$方向的这条边就在最短路上，应该纳入统计。这个题可能会卡SPFA，所以使用了堆优化的DIJ

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 10005;
const int maxm = 250000 * 2;
struct EDGE
{
    int to, next, len;
    EDGE() {}
    EDGE(int to, int next, int len) : to(to), next(next), len(len) {}
} edge[maxm];
int head[maxn], edgecnt;
int dis[2][maxn];
int n;
struct node
{
    int u, dis;
    node(int u, int dis): u(u), dis(dis) {}
    bool operator < (const node &b) const
    {
        return dis > b.dis;
    }
};
void init()
{
    memset(head, -1, sizeof(head));
    edgecnt = 0;
}
void add(int s, int t, int l)
{
    edge[edgecnt] = EDGE(t, head[s], l);
    head[s] = edgecnt++;
}
void spfa(int st, int dis[])
{
    memset(dis, 0x7F, sizeof(int) *n);
    dis[st] = 0;
    priority_queue<node> q;
    static bool f[maxn];
    memset(f, 0, n);
    q.emplace(st, 0);
    while(!q.empty())
    {
        int u = q.top().u;
        q.pop();
        if(f[u])
            continue;
        f[u] = 1;
        for(int i = head[u]; ~i; i = edge[i].next)
        {
            int v = edge[i].to;
            if(dis[v] > dis[u] + edge[i].len)
            {
                dis[v] = dis[u] + edge[i].len;
                q.emplace(v, dis[v]);
            }
        }
    }
}
int main()
{
    int m;
    while(~scanf("%d %d", &n, &m))
    {
        init();
        for(int i = 1; i <= m; i++)
        {
            int s, t, l;
            scanf("%d %d %d", &s, &t, &l);
            add(s, t, l);
            add(t, s, l);
        }
        spfa(0, dis[0]);
        spfa(n - 1, dis[1]);
        int len = dis[0][n - 1];
        int ans = 0;
        for(int u = 0; u < n; u++)
        {
            for(int i = head[u]; ~i; i = edge[i].next)
            {
                int v = edge[i].to;
                if(v == u)
                    continue;
                if(dis[0][u] + edge[i].len + dis[1][v] == len)
                    ans += edge[i].len;
            }
        }
        printf("%d\n", 2 * ans);
    }
    return 0;
}

C - Golf Bot

使用FFT进行计数。使用$cnt$数组记录每个值是否出现。将$cnt$数组和自己进行卷积后（假设为$A$数组），$A[i]$的意义就是任意选两个存在的数进行求和后，和值为$i$的方案数。再将$A$数组加回到$cnt$数组。$cnt[i]$是否为$0$，就代表是否有方案至多选两个数，至少选一个数的和为i。

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#include<bits/stdc++.h>
using namespace std;
const int FFT_MAXN = 1 << 19;
const double pi = acos(-1.0);
struct Complex
{
    double R, I;
    inline Complex(double real = 0.0, double image = 0.0)
    {
        R = real, I = image;
    }
    inline Complex operator + (Complex const &b) const
    {
        return Complex(R + b.R, I + b.I);
    }
    inline Complex operator - (Complex const &b) const
    {
        return Complex(R - b.R, I - b.I);
    }
    inline Complex operator * (Complex const &b) const
    {
        return Complex(R * b.R - I * b.I, I * b.R + R * b.I);
    }
} Wn[FFT_MAXN + 1];
int bitrev[FFT_MAXN];
inline int getLength(int n, int m)
{
    n = n + m - 1;
    int i = 1;
    for(; i < n; i <<= 1);
    return i;
}
void Change(Complex a[], int n)
{
    int d = 0;
    while((1 << d)*n != FFT_MAXN)
        d++;
    for(int i = 0; i < n; i++)
        if(i < (bitrev[i] >> d))
            swap(a[i], a[bitrev[i] >> d]);
}
void FFT(Complex P[], int n, int op)
{
    Change(P, n);
    for(int len = 2; len <= n; len <<= 1)
    {
        int m = len >> 1;
        int del = FFT_MAXN / len * op;
        for(int i = 0; i < n; i += len)
        {
            Complex *p = P + i, *q = P + i + m, *w = op == 1 ? Wn : Wn + FFT_MAXN;
            for(int j = 0; j < m; j++, p++, q++, w += del)
            {
                Complex ne = *q * *w;
                *q = *p - ne;
                *p = *p + ne;
            }
        }
    }
    if(op == -1)
        for(int i = 0; i < n; i++)
        {
            P[i].R /= n;
            P[i].I /= n;
        }
}
void FFTInit()
{
    int L = 0;
    while((1 << L) != FFT_MAXN)
        L++;
    bitrev[0] = 0;
    for(int i = 1; i < FFT_MAXN; i++)
        bitrev[i] = bitrev[i >> 1] >> 1 | ((i & 1) << (L - 1));
    for(int i = 0; i <= FFT_MAXN; i++)
        Wn[i] = Complex(cos(2 * pi / FFT_MAXN * i), sin(2 * pi / FFT_MAXN * i));
}
Complex A[FFT_MAXN], B[FFT_MAXN];
long long cnt[FFT_MAXN];
int main()
{
    FFTInit();
    int n;
    while(~scanf("%d", &n))
    {
        memset(cnt, 0, sizeof(cnt));
        for(int i = 1; i <= n; i++)
        {
            int x;
            scanf("%d", &x);
            cnt[x]++;
        }
        for(int i = 0; i < FFT_MAXN; i++)
            A[i] = Complex(cnt[i], 0);
        FFT(A, FFT_MAXN, 1);
        for(int i = 0; i < FFT_MAXN; i++)
            A[i] = A[i] * A[i];
        FFT(A, FFT_MAXN, -1);
        for(int i = 0; i < FFT_MAXN; i++)
        {
            int t = A[i].R + 0.5;
            cnt[i] += t;
        }
        int m;
        scanf("%d", &m);
        int ans = 0;
        while(m--)
        {
            int x;
            scanf("%d", &x);
            if(cnt[x])
                ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}

D - Book Club

最大流求是否有合法方案

给出一些给书的关系，问是否有方案，使得每个人可以交出一本书，并且获得一本书，不能自己和自己玩。

将一个人$i$看作两个点，编号$A_i$和$B_i$。从源点引边到每一个$A_i$点，容量为1，表示$i$这个人要交出一本书。从每一个$B_i$引一条边到汇点，容量为1，表示$i$这个人要获得一本书。如果有一个关系，$i$的书可以给$j$，那么连一条边从$A_i$到$B_j$。最后跑最大流，如果流量为n，就满足要求。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5;
const int maxm = 2e5;
struct EDGE
{
    int to, next, flow;
    EDGE(int to = 0, int next = 0, int flow = 0): to(to), next(next), flow(flow) {}
} edge[maxm];
int head[maxn], edgecnt;
void init()
{
    memset(head, -1, sizeof(head));
    edgecnt = 0;
}
void AddEdge(int s, int t, int c)
{
    edge[edgecnt] = EDGE(t, head[s], c);
    head[s] = edgecnt++;
    edge[edgecnt] = EDGE(s, head[t], 0);
    head[t] = edgecnt++;
}
int layer[maxn];
int p[maxn];
bool GetLayer(int st, int en)
{
    memset(layer, 0, sizeof(int) * (en + 1));
    layer[st] = 1;
    queue<int>q;
    q.push(st);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for (int i = head[u]; ~i; i = edge[i].next)
        {
            if (edge[i].flow == 0) continue;
            int v = edge[i].to;
            if (layer[v]) continue;
            layer[v] = layer[u] + 1;
            q.push(v);
        }
    }
    return layer[en];
}
int dfsFlow(int u, int en, int f)
{
    int ret = 0;
    if (u == en) return f;
    if (layer[u] >= layer[en]) return 0;
    for (; ~p[u]; p[u] = edge[p[u]].next)
    {
        if (edge[p[u]].flow == 0) continue;
        int v = edge[p[u]].to;
        if (layer[v] != layer[u] + 1) continue;
        int temp = dfsFlow(v, en, min(f, edge[p[u]].flow));
        ret += temp;
        f -= temp;
        edge[p[u]].flow -= temp;
        edge[p[u] ^ 1].flow += temp;
        if (f == 0) break;
    }
    return ret;
}
int dinic(int st, int en)
{
    int ans = 0;
    while (GetLayer(st, en))
    {
        memcpy(p, head, sizeof(int) * (en + 1));
        ans += dfsFlow(st, en, INT_MAX);
    }
    return ans;
}
int main()
{
    init();
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        AddEdge(0, i, 1);
        AddEdge(n + i, n + n + 1, 1);
    }
    while (m--)
    {
        int u, v;
        scanf("%d %d", &u, &v);
        u++;
        v++;
        AddEdge(u, n + v, 1);
    }
    int flow = dinic(0, n + n + 1);
    if (flow != n)
        puts("NO");
    else puts("YES");
    return 0;
}

G - Playing With Geometry

离散化，坐标旋转。

给定两个图，以转角坐标的方式逆时针给出，问两个图的轮廓是否相似。允许进行压缩或者旋转。

直接将两个图的横纵坐标进行离散化（压缩）。枚举将第一图进行旋转，和第二个图进行匹配，匹配的时候，需要找到两张图最左下角的点，让后从这个点开始，逆时针匹配。如果每个坐标都是相同的，那么就是OK的。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 505;
int n;
void Compress(pair<int, int> p[maxn])
{
    int x[maxn], y[maxn];
    for (int i = 0; i < n; i++)
    {
        x[i] = p[i].first;
        y[i] = p[i].second;
    }
    sort(x, x + n);
    sort(y, y + n);
    int cntx = unique(x, x + n) - x - 1;
    int cnty = unique(y, y + n) - y - 1;
    for (int i = 0; i < n; i++)
    {
        p[i].first = lower_bound(x, x + cntx, p[i].first) - x;
        p[i].second = lower_bound(y, y + cnty, p[i].second) - y;
    }
}
void Rotate(pair<int, int> p[maxn])
{
    int ym = 0;
    for (int i = 0; i < n; i++) ym = max(ym, p[i].second);
    for (int i = 0; i < n; i++)
        p[i] = make_pair(ym - p[i].second, p[i].first);
}
bool Compare(pair<int, int> p[2][maxn])
{
    int p1 = min_element(p[0], p[0] + n) - p[0];
    int p2 = min_element(p[1], p[1] + n) - p[1];
    for (int i = 0; i < n; i++)
        if (p[0][(p1 + i) % n] != p[1][(p2 + i) % n]) return false;
    return true;
}
bool check(pair<int, int> p[2][maxn])
{
    for (int i = 0; i < 4; i++)
    {
        if (Compare(p)) return true;
        Rotate(p[0]);
    }
    return false;
}
int main()
{
    pair<int, int> p[2][maxn];
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d %d", &p[0][i].first, &p[0][i].second);
    int m;
    scanf("%d", &m);
    if (n != m)
    {
        puts("no");
        return 0;
    }
    for (int i = 0; i < n; i++)
        scanf("%d %d", &p[1][i].first, &p[1][i].second);
    Compress(p[0]);
    Compress(p[1]);
    if (check(p))
        puts("yes");
    else
        puts("no");
    return 0;
}

I - The Safe Secret

循环枚举，区间DP

给一个由数字和运算符号组成的序列，长度为$2*n$运算符号只包含+-*和?,?可以任意替换为+-*。枚举将序列向左进行循环位移，每次位移$2$个单位，可以得到$n$个表达式序列，问这$n$个表达式的值的最小值和最大值的绝对值。

先将原始序列的数字和符号进行分离。再分别复制一份后接到后面。这样就可以进行区间dp了。注意，第i个数字就是$num[i]$了，它后面的符号就是$symbol[i]$。用$mx[i][j]$来表示从第$i$个数字到第$j$个数字组成的子区间的表达式的最大值，$mi[i][j]$表示最小值。初始化$mx[i][i] = mi[i][i] = num[i]$。枚举区间长度，枚举区间起点，再枚举区间内的断点，进行转移即可。要注意断点处为乘号的时候，并不是最小值*最小值就是最小值（考虑两个都是负数的情况），为了偷懒，直接暴力枚举了4种情况进行转移。

最后mx[i][n+i-1]和mi[i][n+i-1]就是第i个序列的表达式的最大值和最小值

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#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 405;
LL mx[maxn][maxn];
LL mi[maxn][maxn];
void getMaxMin(const vector<int> &num, const vector<char> &symbol)
{
    memset(mx, 0x80, sizeof(mx));
    memset(mi, 0x7F, sizeof(mi));
    int n = num.size();
    for (int i = 0; i < n; i++)
        mi[i][i] = mx[i][i] = num[i];
    for (int len = 2; len <= n / 2; len++)
    {
        for (int st = 0; st + len <= n; st++)
        {
            int en = st + len - 1;
            for (int mid = st; mid < en; mid++)
            {
                if (symbol[mid] == '+' || symbol[mid] == '?')
                {
                    mi[st][en] = min(mi[st][en], mi[st][mid] + mi[mid + 1][en]);
                    mx[st][en] = max(mx[st][en], mx[st][mid] + mx[mid + 1][en]);
                }
                if (symbol[mid] == '-' || symbol[mid] == '?')
                {
                    mi[st][en] = min(mi[st][en], mi[st][mid] - mx[mid + 1][en]);
                    mx[st][en] = max(mx[st][en], mx[st][mid] - mi[mid + 1][en]);
                }
                if (symbol[mid] == '*' || symbol[mid] == '?')
                {
                    mi[st][en] = min(mi[st][en], mi[st][mid] * mi[mid + 1][en]);
                    mi[st][en] = min(mi[st][en], mi[st][mid] * mx[mid + 1][en]);
                    mi[st][en] = min(mi[st][en], mx[st][mid] * mi[mid + 1][en]);
                    mi[st][en] = min(mi[st][en], mx[st][mid] * mx[mid + 1][en]);
                    mx[st][en] = max(mx[st][en], mx[st][mid] * mx[mid + 1][en]);
                    mx[st][en] = max(mx[st][en], mx[st][mid] * mi[mid + 1][en]);
                    mx[st][en] = max(mx[st][en], mi[st][mid] * mx[mid + 1][en]);
                    mx[st][en] = max(mx[st][en], mi[st][mid] * mi[mid + 1][en]);
                }
            }
        }
    }
}
int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        vector<int> num;
        vector<char> symbol;
        for (int i = 0; i < n; i++)
        {
            int x;
            scanf("%d", &x);
            num.push_back(x);
            char s[10];
            scanf("%s", s);
            symbol.push_back(s[0]);
        }
        for (int i = 0; i < n; i++)
        {
            num.push_back(num[i]);
            symbol.push_back(symbol[i]);
        }
        getMaxMin(num, symbol);
        for (int i = 0; i < n; i++)
            printf("%lld%lld", abs(mi[i][i + n - 1]), abs(mx[i][i + n - 1]));
        printf("\n");
    }
    return 0;
}