HDU 4289：Control（最小割）

http://acm.hdu.edu.cn/showproblem.php?pid=4289

题意：有n个城市，m条无向边，小偷要从s点开始逃到d点，在每个城市安放监控的花费是sa[i]，问最小花费可以监控到所有小偷。

思路：求最小割可以转化为最大流。每个城市之间拆点，流量是sa[i]，再增加一个超级源点S和s相连，增加一个超级汇点T，让d的第二个点和T相连。然后就可以做了。

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <vector>
 #include <map>
 #include <set>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define N 100010
 #define M 510
 typedef long long LL;
 struct Edge {
     int v, nxt, cap;
     Edge () {}
     Edge (int v, int cap, int nxt) : v(v), cap(cap), nxt(nxt) {}
 }edge[N*];
 int head[M], tot, gap[M], dis[M], cur[M], pre[M];
 int sa[M], S, T;

 void Add(int u, int v, int cap) {
     edge[tot] = Edge(v, cap, head[u]); head[u] = tot++;
     edge[tot] = Edge(u, , head[v]); head[v] = tot++;
 }

 void BFS() {
     memset(dis, -, sizeof(dis));
     memset(gap, , sizeof(gap));
     queue<int> que; que.push(T);
     dis[T] = ; gap[]++;
     while(!que.empty()) {
         int u = que.front(); que.pop();
         for(int i = head[u]; ~i; i = edge[i].nxt) {
             Edge& e = edge[i];
             if(~dis[e.v]) continue;
             dis[e.v] = dis[u] + ;
             gap[dis[e.v]]++;
             que.push(e.v);
         }
     }
 }

 int ISAP(int n) {
     BFS(); // 别忘了调用!
     memcpy(cur, head, sizeof(cur));
     int u = pre[S] = S, ans = , i;
     while(dis[S] < n) {
         if(u == T) {
             int flow = INF, index;
             for(i = S; i != T; i = edge[cur[i]].v)
                 if(flow > edge[cur[i]].cap)
                     flow = edge[cur[i]].cap, index = i;
             for(i = S; i != T; i = edge[cur[i]].v)
                 edge[cur[i]].cap -= flow, edge[cur[i]^].cap += flow;
             u = index; ans += flow;
         }
         for(i = cur[u]; ~i; i = edge[i].nxt)
             if(dis[edge[i].v] == dis[u] -  && edge[i].cap > ) break;
         if(~i) {
             pre[edge[i].v] = u; cur[u] = i;
             u = edge[i].v;
         } else {
             if(--gap[dis[u]] == ) break;
             int md = n;
             for(i = head[u]; ~i; i = edge[i].nxt)
                 if(edge[i].cap >  && dis[edge[i].v] < md)
                     md = dis[edge[i].v], cur[u] = i;
             ++gap[dis[u] = md + ];
             u = pre[u];
         }
     }
     return ans;
 }

 int main() {
     int n, m;
     while(~scanf("%d%d", &n, &m)) {
         int s, t;
         scanf("%d%d", &s, &t);
         memset(head, -, sizeof(head)); tot = ;
         S = , T =  * n + ;
         // 第一个点是1~n,第二个点是n+1~2*n,第一个点为进来的点,第二个点为出去的点
         Add(S, s, INF); Add(t + n, T, INF);
         for(int i = ; i <= n; i++) scanf("%d", &sa[i]);
         for(int i = ; i <= m; i++) {
             int u, v;
             scanf("%d%d", &u, &v);
             Add(u + n, v, INF);
             Add(v + n, u, INF);
         }
         for(int i = ; i <= n; i++)
             Add(i, i + n, sa[i]);
         int ans = ISAP(T + );
         printf("%d\n", ans);
     }
     return ;
 }