POJ--1273--Drainage Ditches【Dinic】网络最大流

链接：http://poj.org/problem?id=1273

题意：农夫的农场被水淹了，他建了一些排水沟来排水，最终把这些水排到小河里，现有n个点，节点1~n-1为池塘，水从1开始流，n为小河。然后有m条排水沟，每条排水沟告诉起点、终点、最大水流速度，现在求这个排水系统的最大排水速度。

网络最大流裸题，dinic模板

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 100100
#define eps 1e-7
#define INF 0x7FFFFFFF
#define seed 131
#define ll long long
#define ull unsigned ll
#define lson l,m,rt<<1
#define rson r,m+1,rt<<1|1

struct node{
    int u,w,next;
}edge[5000];
int head[2100],vis[2100],dist[2100];
int n,m,src,sink,cnt;
void add_edge(int a,int b,int c){
    edge[cnt].u = b;
    edge[cnt].w = c;
    edge[cnt].next = head[a];
    head[a] = cnt++;
}
void bfs(){
    int i, j;
    memset(dist,0,sizeof(dist));
    queue<int>q;
    vis[src] = 1;
    q.push(src);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(i=head[u];i!=-1;i=edge[i].next){
            if(!vis[edge[i].u]&&edge[i].w){
                q.push(edge[i].u);
                dist[edge[i].u] = dist[u] + 1;
                vis[edge[i].u] = 1;
            }
        }
    }
}
int dfs(int u,int delta){
    int i,j;
    if(u==sink) return delta;
    else{
        int ret = 0;
        for(i=head[u];i!=-1&δi=edge[i].next){
            if(edge[i].w&&dist[edge[i].u]==dist[u]+1){
                int dd = dfs(edge[i].u,min(edge[i].w,delta));
                edge[i].w -= dd;
                edge[i^1].w += dd;
                delta -= dd;
                ret += dd;
            }
        }
        return ret;
    }
}
int maxflow(){
    int ret = 0;
    while(1){
        memset(vis,0,sizeof(vis));
        bfs();
        if(!vis[sink])  break;
        ret += dfs(src,INF);
    }
    return ret;
}
int main(){
    int i,j;
    int a,b,x;
    while(scanf("%d%d",&m,&n)!=EOF){
        memset(head,-1,sizeof(head));
        int cnt = 0;
        src = 1;
        sink = n;
        for(i=0;i<m;i++){
            scanf("%d%d%d",&a,&b,&x);
            add_edge(a,b,x);
            add_edge(b,a,0);
        }
        int ans = maxflow();
        printf("%d\n",ans);
    }
    return 0;
}

POJ--1273--Drainage Ditches【Dinic】网络最大流