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 = 205;
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=0,int _c=0,int _nex=0):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()-1;}

bool bfs() {
    queue<int> q;
    CLR(dep, -1);
    q.push(S); dep[S] = 0;
    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] == -1) {
                dep[v] = dep[u] + 1;
                q.push(v);
            }
        }
    }
    return dep[T] != -1;
}
int dfs(int u, int flow) {
    if(u == T) return flow;
    int w, used=0;
    for(int i = head[u]; ~i; i = E[i].nex) {
        int v = E[i].v;
        if(dep[v] == dep[u] + 1) {
            w = flow - used;
            w = dfs(v, min(w, E[i].c));
            E[i].c -= w;  E[i^1].c += w;
            if(v) cur[u] = i;
            used += w;
            if(used == flow) return flow;
        }
    }
    if(!used) dep[u] = -1;
    return used;
}
int dinic() {
    int ans = 0;
    while(bfs()) {
        for(int i = 1; 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, -1);
        scanf("%d%d", &n, &m);
        scanf("%d%d", &S, &T);
        for(i = 1; i <= m; ++i) {
            scanf("%d%d%d", &u, &v, &w);
            add(u, v, w*300+1); add(v, u, 0);
        }
        int ans = dinic()%300;
        printf("%d
", ans);
    }
    return 0;
}