HDU 4289 Control 最小割

Control

题意：有一个犯罪集团要贩卖大规模杀伤武器，从s城运输到t城，现在你是一个特殊部门的长官，可以在城市中布置眼线，但是布施眼线需要花钱，现在问至少要花费多少能使得你及时阻止他们的运输。

题解：裸的最小割模型，最小割就是最大流，我们把点拆成2个点，然后将原点与拆点建边，流量为在城市建立眼线的费用，然后拆点为出点，原点为入点，将可以到达的城市之间建流量为无穷的边。

最后求出s 到 t的拆点的最大流 那么就是这个题目的答案了。

代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb emplace_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define lch(x) tr[x].son[0]
 #define rch(x) tr[x].son[1]
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const int inf = 0x3f3f3f3f;
 const LL INF = 0x3f3f3f3f3f3f3f3f;
 const LL mod =  (int)1e9+;
 const int N = ;
 const int M = N*N*;
 int head[N], deep[N], cur[N];
 int w[M], to[M], nx[M];
 int tot;
 int n, m, s, t;
 int u, v, val;
 void add(int u, int v, int val){
     w[tot]  = val; to[tot] = v;
     nx[tot] = head[u]; head[u] = tot++;

     w[tot] = ; to[tot] = u;
     nx[tot] = head[v]; head[v] = tot++;
 }
 int bfs(int s, int t){
     queue<int> q;
     memset(deep, , sizeof(deep));
     q.push(s);
     deep[s] = ;
     while(!q.empty()){
         int u = q.front();
         q.pop();
         for(int i = head[u]; ~i; i = nx[i]){
             if(w[i] >  && deep[to[i]] == ){
                 deep[to[i]] = deep[u] + ;
                 q.push(to[i]);
             }
         }
     }
     return deep[t] > ;
 }
 int Dfs(int u, int t, int flow){
     if(u == t) return flow;
     for(int &i = cur[u]; ~i; i = nx[i]){
         if(deep[u]+ == deep[to[i]] && w[i] > ){
             int di = Dfs(to[i], t, min(w[i], flow));
             if(di > ){
                 w[i] -= di, w[i^] += di;
                 return di;
             }
         }
     }
     return ;
 }

 int Dinic(int s, int t){
     int ans = , tmp;
     while(bfs(s, t)){
         for(int i = ; i <= *n+; i++) cur[i] = head[i];
         while(tmp = Dfs(s, t, inf)) ans += tmp;
     }
     return ans;
 }
 void init(){
     memset(head, -, sizeof(head));
     tot = ;
 }
 int main(){
     while(~scanf("%d%d", &n, &m)){
         init();
         int ss = n*+, tt = ss+;
         s = ss, t = tt;
         scanf("%d%d", &ss, &tt);
         add(s, ss, inf);
         add(tt+n, t, inf);
         for(int i = ; i <= n; i++){
             scanf("%d", &val);
             add(i,i+n,val);
         }
         for(int i = ; i <= m; i++){
             scanf("%d%d",&u,&v);
             add(u+n,v,inf);
             add(v+n,u,inf);
         }
         printf("%d\n",Dinic(s,t));
     }
     return ;
 }