hdu 6214 Smallest Minimum Cut[最大流]

hdu 6214 Smallest Minimum Cut[最大流]

题意：求最小割中最少的边数。

题解：对边权乘个比边大点的数比如300，再加1 ，最后，最大流对300取余就是边数啦。。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<vector>
 #include<queue>
 using namespace std;
 #define CLR(a,b) memset((a),(b),sizeof((a)))
 typedef long long ll;
 const int N = ;
 const int inf = 0x3f3f3f3f;
 int n, m, S, T;
 int dep[N], cur[N];
 int head[N];
 struct Edge{
     int v, c, nex;
     Edge(int _v=,int _c=,int _nex=):v(_v),c(_c),nex(_nex){}
 };
 vector<Edge>E;
 void add(int u,int v,int c){E.push_back(Edge(v,c,head[u]));head[u]=E.size()-;}

 bool bfs() {
     queue<int> q;
     CLR(dep, -);
     q.push(S); dep[S] = ;
     while(!q.empty()) {
         int u = q.front(); q.pop();
         for(int i = head[u]; ~i; i = E[i].nex) {
             int v = E[i].v;
             if(E[i].c && dep[v] == -) {
                 dep[v] = dep[u] + ;
                 q.push(v);
             }
         }
     }
     return dep[T] != -;
 }
 int dfs(int u, int flow) {
     if(u == T) return flow;
     int w, used=;
     for(int i = head[u]; ~i; i = E[i].nex) {
         int v = E[i].v;
         if(dep[v] == dep[u] + ) {
             w = flow - used;
             w = dfs(v, min(w, E[i].c));
             E[i].c -= w;  E[i^].c += w;
             if(v) cur[u] = i;
             used += w;
             if(used == flow) return flow;
         }
     }
     if(!used) dep[u] = -;
     return used;
 }
 int dinic() {
     int ans = ;
     while(bfs()) {
         for(int i = ; i <= T;i++)
             cur[i] = head[i];
         ans += dfs(S, inf);
     }
     return ans;
 }
 int main() {
     int t, i, u, v, w;
     scanf("%d", &t);
     while(t--) {
         E.clear(); CLR(head, -);
         scanf("%d%d", &n, &m);
         scanf("%d%d", &S, &T);
         for(i = ; i <= m; ++i) {
             scanf("%d%d%d", &u, &v, &w);
             add(u, v, w*+); add(v, u, );
         }
         int ans = dinic()%;
         printf("%d\n", ans);
     }
     return ;
 }

249ms