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。

  1 #include <iostream>
  2 #include <vector>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <algorithm>
  6 #include <cmath>
  7 #include <map>
  8 #include <set>
  9 #include <string>
 10 #include <queue>
 11 using namespace std;
 12 #define pb(x) push_back(x)
 13 #define ll long long
 14 #define mk(x, y) make_pair(x, y)
 15 #define lson l, m, rt<<1
 16 #define mem(a) memset(a, 0, sizeof(a))
 17 #define rson m+1, r, rt<<1|1
 18 #define mem1(a) memset(a, -1, sizeof(a))
 19 #define mem2(a) memset(a, 0x3f, sizeof(a))
 20 #define rep(i, a, n) for(int i = a; i<n; i++)
 21 #define ull unsigned long long
 22 typedef pair<int, int> pll;
 23 const double PI = acos(-1.0);
 24 const double eps = 1e-8;
 25 const int mod = 1e9+7;
 26 const int inf = 1061109567;
 27 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
 28 const int maxn = 4e4+5;
 29 int q[maxn*2], head[maxn*2], dis[maxn/10], s, t, num, edge[2500][2];
 30 struct node
 31 {
 32     int to, nextt, c;
 33     node(){}
 34     node(int to, int nextt, int c):to(to), nextt(nextt), c(c){}
 35 }e[maxn*2];
 36 void init() {
 37     num = 0;
 38     mem1(head);
 39 }
 40 void add(int u, int v, int c) {
 41     e[num] = node(v, head[u], c); head[u] = num++;
 42     e[num] = node(u, head[v], 0); head[v] = num++;
 43 }
 44 int bfs() {
 45     mem(dis);
 46     dis[s] = 1;
 47     int st = 0, ed = 0;
 48     q[ed++] = s;
 49     while(st<ed) {
 50         int u = q[st++];
 51         for(int i = head[u]; ~i; i = e[i].nextt) {
 52             int v = e[i].to;
 53             if(!dis[v]&&e[i].c) {
 54                 dis[v] = dis[u]+1;
 55                 if(v == t)
 56                     return 1;
 57                 q[ed++] = v;
 58             }
 59         }
 60     }
 61     return 0;
 62 }
 63 int dfs(int u, int limit) {
 64     if(u == t) {
 65         return limit;
 66     }
 67     int cost = 0;
 68     for(int i = head[u]; ~i; i = e[i].nextt) {
 69         int v = e[i].to;
 70         if(e[i].c&&dis[v] == dis[u]+1) {
 71             int tmp = dfs(v, min(limit-cost, e[i].c));
 72             if(tmp>0) {
 73                 e[i].c -= tmp;
 74                 e[i^1].c += tmp;
 75                 cost += tmp;
 76                 if(cost == limit)
 77                     break;
 78             } else {
 79                 dis[v] = -1;
 80             }
 81         }
 82     }
 83     return cost;
 84 }
 85 int dinic() {
 86     int ans = 0;
 87     while(bfs()) {
 88         ans += dfs(s, inf);
 89     }
 90     return ans;
 91 }
 92 int main()
 93 {
 94     int n, m, x, y;
 95     while(~scanf("%d%d", &n, &m)) {
 96         for(int i = 0; i<m; i++) {
 97             scanf(" (%d,%d)", &edge[i][0], &edge[i][1]);
 98         }
 99         s = n;
100         int ans = inf;
101         for(int i = 1; i<n; i++) {
102             t = i;
103             init();
104             for(int j = 0; j<m; j++) {
105                 int x = edge[j][0];
106                 int y = edge[j][1];
107                 add(x+n, y, inf);
108                 add(y+n, x, inf);
109             }
110             for(int j = 0; j<n; j++)
111                 add(j, j+n, 1);
112             ans = min(ans, dinic());
113         }
114         if(ans == inf)
115             ans = n;
116         cout<<ans<<endl;
117     }
118     return 0;
119 }