Description:

給定一張無向圖，問至少去掉多少個點， 可以使圖不連通。點數N ≤ 50

思路：先固定一個點s，然後枚舉另一個點t，然後求最少要割掉幾個點使兩點不連通

　　自然聯系到最小割，但最小割是割邊，割點呢？只要把每個點拆成兩個點，割去一個點等價於在網絡中斷開其拆成的兩點中間的邊。

所以將原無向圖中的邊權設為正無窮大，然後將每個點對之間的邊權設為1，然後就能用最小割寫了

#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
using
 
 namespace std;
const int N = 110;
const int M = 10010;
const int INF = 1e9;

int head[N],now;
struct edges{
    int to,next,w;
}edge[M<<1];
void add(int u,int v,int w){ edge[++now] = {v,head[u],w}; head[u] = now;}

int n,m,s,t,dep[N],ans;
void init(){
    memset(edge,0,sizeof(edge)); now = 1
 
;
    memset(head,0,sizeof(head));
    memset(dep,0,sizeof(dep));
}

int dinic(int x,int flow){
    if(x == t) return flow;
    int rest = flow, k;
    for(int i = head[x]; i; i = edge[i].next){
        int v = edge[i].to;
        if(edge[i].w && dep[v] == dep[x] + 1){
            k 
 
= dinic(v,min(rest, edge[i].w));
            if(!k) dep[v] = 0;
            edge[i].w -= k;
            edge[i ^ 1].w += k;
            rest -= k;
        }
    }
    return flow - rest;
}
bool bfs(){
    memset(dep,0,sizeof(dep));
    queue<int> q;
    q.push(s); dep[s] = 1;
    while(!q.empty()){
        int x = q.front(); q.pop();
        for(int i = head[x]; i; i = edge[i].next){
            int v = edge[i].to;
            if(edge[i].w && !dep[v]){
                q.push(v);
                dep[v] = dep[x] + 1;
                if(v == t) return 1;
            }
        }
    }
    return 0;
}
struct input{
    int x,y;
}inp[M];
int main(){
    while(scanf("%d%d",&n,&m) != EOF){
        ans = INF;
        int x,y;
        for(int i = 1; i <= m; i++){
            scanf(" (%d,%d)",&x,&y);
            x++, y++;
            inp[i] = {x,y};
//            add(x,y+n,INF);add(y+n,x,0);
        }
        if(n == 1){
            puts("1");
            continue;
        }
        s = 1;
        for(t = 2; t <= n; t++){  //枚舉一個終點 
            init();
            for(int i = 1; i <= m; i++){
                add(inp[i].y + n,inp[i].x, INF);  //拆點連邊 
                add(inp[i].x,inp[i].y + n, 0);
                add(inp[i].x + n,inp[i].y, INF);
                add(inp[i].y,inp[i].x + n, 0);
            }
            int x,y;
            for(int i = 1; i <= n; i++)
              if(i != s && i != t)
                add(i,i+n,1), add(i+n,i,0);    //把所對拆點之間的邊權設為1
            int tmp = 0, maxflow = 0;
            s += n;
            while(bfs())  //最大流最小割 
              while(tmp = dinic(s,INF)) maxflow += tmp;
            ans = min(ans,maxflow);
            s -= n;
        }
        if(ans == INF) printf("%d\n",n);
        else printf("%d\n",ans);
    }
    return 0;
}

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