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Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14451    Accepted Submission(s): 6840

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

50

题意：n个点，m条单向边，求1到n的最大流。

分析：是一道网络流的基础入门题，也是道模板题。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 1e9;
const int MOD = 1e9+7;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define N 1010

int n,m,s,t;
int pre[N];
int mat[N][N];
bool vis[N];

bool bfs()
{
    int cur;
    queue<int> Q;
    CL(pre, 0);
    CL(vis, false);
    vis[s] = true;
    Q.push(s);
    while(!Q.empty())
    {
        cur = Q.front();
        Q.pop();
        //cout<<cur<<endl;
        if(cur == t) return true;
        for(int i=1; i<=n; i++)
        {
            if(!vis[i] && mat[cur][i])
            {
                Q.push(i);
                pre[i] = cur;
                vis[i] = true;
                //cout<<cur<<"->"<<i<<endl;
            }
        }
    }
    return false;
}

int max_flow()
{
    int ans = 0;
    while(1)
    {

        if(!bfs()) return ans;
        int Min = INF;
        for(int i=t; i!=s; i=pre[i])
            Min = min(Min, mat[pre[i]][i]);
        for(int i=t; i!=s; i=pre[i])
        {
            mat[pre[i]][i] -= Min;
            mat[i][pre[i]] += Min;
        }
        ans += Min;
    }
}

int main()
{
    int u,v,c;
    while(scanf("%d%d",&m,&n)==2)
    {
        CL(mat, 0);
        s = 1;
        t = n;
        //Q.push(s);
        while(m--)
        {
            scanf("%d%d%d",&u,&v,&c);
            mat[u][v] += c;
        }
        printf("%d\n",max_flow());
    }
    return 0;
}

HDU1532 Drainage Ditches （网络流）

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原文地址：http://blog.csdn.net/d_x_d/article/details/51733466