hdu 4289 Control 网络流

题目链接

给出一些点， 每个点有一个权值， 给出一些边， 起点以及终点， 去掉一些点使得起点和终点不连通， 求最小的val。

拆点， 把一个点s拆成s和s'， 之间建一条边， 权值为点权。 对于一条边， <u, v> 建边<u, v'>, <v, u'> 权值为inf， 跑一遍最大流。

 #include<bits/stdc++.h>
 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 maxx = ;
 const int maxn = ;
 int head[maxn*], s, t, num, q[maxn*], dis[maxn];
 struct node
 {
     int to, nextt, c;
 }e[maxn*];
 void init() {
     mem1(head);
     num = ;
 }
 void add(int u, int v, int c) {
     e[num].to = v; e[num].nextt = head[u]; e[num].c = c; head[u] = num++;
     e[num].to = u; e[num].nextt = head[v]; e[num].c = ; head[v] = num++;
 }
 int bfs() {
     int u, v, st = , ed = ;
     mem(dis);
     dis[s] = ;
     q[ed++] = s;
     while(st<ed) {
         u = q[st++];
         for(int i = head[u]; ~i; i = e[i].nextt) {
             v = e[i].to;
             if(e[i].c&&!dis[v]) {
                 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[u] == dis[v]-) {
             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, val, x, y;
     while(cin>>n>>m) {
         cin>>s>>t;
         t+=n;
         init();
         for(int i = ; i<=n; i++) {
             scanf("%d", &val);
             add(i, i+n, val);
         }
         while(m--) {
             scanf("%d%d", &x, &y);
             add(x+n, y, inf);
             add(y+n, x, inf);
         }
         int ans = dinic();
         cout<<ans<<endl;
     }
 }