POJ1273 Drainage Ditches （最大流+入门题）

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<climits>
#include<cstring>
#include<queue>
#include<vector>
#include<map>
#include<set>
using namespace std;

int n,m;
int cap[210][210];
int flow[210][210];
int a[210];
const int INF=1000000000;

int EK(int s,int t)
{
    queue<int> q;
    memset(flow,0,sizeof(flow));
    int p[210];
    int f=0;
    for(;;)
    {
        memset(a,0,sizeof(a));
        a[s]=INF;
        q.push(s);
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            for(int v=1;v<=m;v++)if(!a[v]&&cap[u][v]>flow[u][v])
            {
                p[v]=u;
                q.push(v);
                a[v]=min(a[u],cap[u][v]-flow[u][v]);
            }
        }
        if(a[t]==0)break;
        for(int u=t;u!=s;u=p[u])
        {
            flow[p[u]][u]+=a[t];
            flow[u][p[u]]-=a[t];
        }
        f+=a[t];
    }
    return f;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(cap,0,sizeof(cap));
        for(int i=0;i<n;i++)
        {
            int s,e,c;
            scanf("%d%d%d",&s,&e,&c);
            /*if((cap[s][e]==0)||(cap[s][e]!=0&&c>cap[s][e]))
            {
                cap[s][e]=c;
                //cap[e][s]=0;
            }*/
            cap[s][e]+=c;
        }
        printf("%d\n",EK(1,m));
    }
    return 0;
}