poj 1966 Cable TV Network 顶点连通度

题目链接

给一个图， n个点m条边， 求至少去掉多少个点可以使得图不再联通。
随便指定一个点为源点， 枚举其他点为汇点的情况， 跑网络流， 求其中最小的情况。 如果最后ans为inf， 说明是一个完全图， 那么结果就为n。

 #include <iostream>
 #include <vector>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <map>
 #include <set>
 #include <string>
 #include <queue>
 using namespace std;
 #define pb(x) push_back(x)
 #define ll long long
 #define mk(x, y) make_pair(x, y)
 #define lson l, m, rt<<1
 #define mem(a) memset(a, 0, sizeof(a))
 #define rson m+1, r, rt<<1|1
 #define mem1(a) memset(a, -1, sizeof(a))
 #define mem2(a) memset(a, 0x3f, sizeof(a))
 #define rep(i, a, n) for(int i = a; i<n; i++)
 #define ull unsigned long long
 typedef pair<int, int> pll;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = 4e4+;
 int q[maxn*], head[maxn*], dis[maxn/], s, t, num, edge[][];
 struct node
 {
     int to, nextt, c;
     node(){}
     node(int to, int nextt, int c):to(to), nextt(nextt), c(c){}
 }e[maxn*];
 void init() {
     num = ;
     mem1(head);
 }
 void add(int u, int v, int c) {
     e[num] = node(v, head[u], c); head[u] = num++;
     e[num] = node(u, head[v], ); head[v] = num++;
 }
 int bfs() {
     mem(dis);
     dis[s] = ;
     int st = , ed = ;
     q[ed++] = s;
     while(st<ed) {
         int u = q[st++];
         for(int i = head[u]; ~i; i = e[i].nextt) {
             int v = e[i].to;
             if(!dis[v]&&e[i].c) {
                 dis[v] = dis[u]+;
                 if(v == t)
                     return ;
                 q[ed++] = v;
             }
         }
     }
     return ;
 }
 int dfs(int u, int limit) {
     if(u == t) {
         return limit;
     }
     int cost = ;
     for(int i = head[u]; ~i; i = e[i].nextt) {
         int v = e[i].to;
         if(e[i].c&&dis[v] == dis[u]+) {
             int tmp = dfs(v, min(limit-cost, e[i].c));
             if(tmp>) {
                 e[i].c -= tmp;
                 e[i^].c += tmp;
                 cost += tmp;
                 if(cost == limit)
                     break;
             } else {
                 dis[v] = -;
             }
         }
     }
     return cost;
 }
 int dinic() {
     int ans = ;
     while(bfs()) {
         ans += dfs(s, inf);
     }
     return ans;
 }
 int main()
 {
     int n, m, x, y;
     while(~scanf("%d%d", &n, &m)) {
         for(int i = ; i<m; i++) {
             scanf(" (%d,%d)", &edge[i][], &edge[i][]);
         }
         s = n;
         int ans = inf;
         for(int i = ; i<n; i++) {
             t = i;
             init();
             for(int j = ; j<m; j++) {
                 int x = edge[j][];
                 int y = edge[j][];
                 add(x+n, y, inf);
                 add(y+n, x, inf);
             }
             for(int j = ; j<n; j++)
                 add(j, j+n, );
             ans = min(ans, dinic());
         }
         if(ans == inf)
             ans = n;
         cout<<ans<<endl;
     }
     return ;
 }