HDU 1532 Drainage Ditches 最大流入门

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本文介绍了一个典型的最大流问题实例——排水沟渠问题。通过构建复杂网络来确定从池塘到溪流的最大排水量，利用Edmonds-Karp算法寻找并增强瓶颈路径以实现最大流。文章提供了完整的C++代码实现。

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Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5875 Accepted Submission(s): 2779

Problem Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

Sample Output

Source

USACO 93

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lwg

最大流第一题，题意是：输入N，M。N代表沟渠的数量。M代表沟渠的路分口的数量。紧接着又N条沟渠。每行输入S,E,C。代表从S到E最多能流C的水，求从1到M最多能排出多少水。

思路：就是从S到T找出每条路最小的残量，然后将路径上的流量加上这个残量，直到残量为0的时候，即为最大流。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define INF 9999999
int n,m;
int map[300][300],pre[300],flow[300][300],p[300],a[300];
int EK(int s,int t)
{
    int sum=0;
    queue<int>q;
    memset(flow,0,sizeof(flow));
    for(;;)
    {
        memset(a,0,sizeof(a));//记录残量
        a[s]=INF;
        q.push(s);
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            for(int i=2; i<=m; i++)
                if(!a[i]&&map[u][i]>flow[u][i])
                {
                    p[i]=u;//记录i的父亲节的是u
                    q.push(i);
                    a[i]=a[u]<map[u][i]-flow[u][i]?a[u]:map[u][i]-flow[u][i];
                }
        }
        if(!a[t])break;//如果残量是0的话，就找到最大流
        for(int i=t; i!=s; i=p[i])//每条路加上最小残量
        {
            flow[p[i]][i]+=a[t];
            flow[i][p[i]]-=a[t];
        }
        sum+=a[t];//记录流量
    }
    return sum;
}
int main()
{
    int a,b,c,max;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(map,0,sizeof(map));
        memset(p,0,sizeof(p));
        
        while(n--)
        {
            scanf("%d%d%d",&a,&b,&c);
            map[a][b]+=c;
        }
        max=EK(1,m);
        printf("%d\n",max);
    }

    return 0;
}