Drainage Ditches （HDU - 1532）（最大流）

HDU - 1532

题意：有m个点，n条管道，问从1到m最大能够同时通过的水量是多少？

题解：最大流模板题。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 0x3ffffff;
const int maxn = 205;
int gra[maxn][maxn];
int flow[maxn];
int path[maxn];
int n,m;

int bfs(int s, int e)
{
    queue<int>q;
    int t;
    memset(path,-1,sizeof(path));
    memset(flow,0,sizeof(flow));
    path[s] = 0;
    flow[s] = INF;
    q.push(s);
    while(!q.empty()){
        t = q.front();
        q.pop();
        if(t == e) break;
        for(int i = 1; i <= m; i ++)
        {
            if(i != s && gra[t][i] && path[i] == -1)
            {
                flow[i] = flow[t] < gra[t][i] ? flow[t] : gra[t][i];
                path[i] = t;
                q.push(i);
            }
        }
    }
    if(path[e] == -1)
        return -1;
    else return flow[e];
}

int EK(int s, int e){
    int maxflow = 0;
    int pre, now, step;
    while((step = bfs(s,e))!= -1){
        maxflow += step;
        now = e;
        while(now != s){
            pre = path[now];
            gra[pre][now] -= step;
            gra[now][pre] += step;
            now = pre;
        }
    }
    return maxflow;
}
int main()
{

    int u,v,c;
    while(~scanf("%d%d",&n,&m)){
        memset(gra,0,sizeof(gra));
        for(int i = 1; i <= n; i ++)
        {
            scanf("%d %d %d", &u,&v,&c);
            gra[u][v] += c;
        }
        int ans = EK(1,m);
        printf("%d\n",ans);
    }
    return 0;
}