Drainage Ditches

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

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

              

Sample Output

 50 

 

用的EK，很简单的一个最大流模板题，看我的注释吧。。。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
#define oo 1<<28
using namespace std;
int n,m,map[300][300];
int a[300],p[300],flow[300][300];
int start,end;
int Dinic()
{
    int u,v;
    queue<int>q;
    memset(flow,0,sizeof(flow));
    memset(p,0,sizeof(p));
    int f=0;
    while(1)
    {
        memset(a,0,sizeof(a));
        a[start]=oo;//oo是因为起点没有节点控制，先初始为最
        q.push(start);
        while(!q.empty())//bfs找增广路
        {
            int u=q.front();
            q.pop();
            for(v=1;v<=m;v++)
            {
                if(!a[v]&&map[u][v]>flow[u][v])//找到新节点，保证要流过的小于最大界限
                {
                    p[v]=u;
                    q.push(v);
                    a[v]=a[u]<map[u][v]-flow[u][v]?a[u]:map[u][v]-flow[u][v];//最小残量，保证此增光路上每一条都可通过，否则就成了死路
                }
            }
        }
        if(a[end]==0)
            break;
        for(u=end;u!=start;u=p[u])//从汇点向回流
        {
            flow[p[u]][u]+=a[end];//正向的加上此增广路上最小的值
            flow[u][p[u]]-=a[end];//同样反向的要减去
        }
        f+=a[end];
    }
    return f;
}
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        memset(map,0,sizeof(map));
        memset(flow,0,sizeof(flow));
        int u,v,w;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            map[u][v]+=w; //防止有重边
        }
        start=1,  end=m;//开始于1，结束于m
        printf("%d\n",Dinic());
    }
   return 0;
}