poj 1273 Drainage Ditches 最大流问题.

Drainage Ditches 原文链接 click here.

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 68476 Accepted: 26518
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
Source

USACO 93

这是模板题,直接套模板就可以了,

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<queue>
#include<math.h>p
using namespace std;
const int N = 201;
const int INF = 0x3f3f3f;
int n,m,sum,s,t;//s,t分别为源点和终点;sum为最大流量;
int flow[N][N],cap[N][N],a[N],p[N];
//分别表示流量,容量,当前最小的残余量, 之前参残余量;
void karp()
{
    int i,u,v;
    queue<int >q;
    while(1)
    {
        memset(a,0,sizeof(a));
        a[s]=INF;
        q.push(s);
        while(!q.empty())
        {
            u=q.front();
            q.pop();
            for(v=1; v<=m; v++)
            {
                if(!a[v]&&flow[u][v]<cap[u][v])
                {
                    p[v]=u;
                    q.push(v);
                    a[v]=min(a[u],cap[u][v]-flow[u][v]);
                }
            }
        }
        if(a[m]==0)
            break;
        sum+=a[m];
        for(i=m; i!=s; i=p[i])
        {
            flow[p[i]][i]+=a[m];
            flow[i][p[i]]-=a[m];
        }
    }
    printf("%d\n",sum);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int a,b,c;
        s=1,t=m,sum=0;
        //s 一直都是1;
        memset(flow,0,sizeof(flow));
        memset(cap,0,sizeof(cap));
        while(n--)// n 条路;
        {
            scanf("%d%d%d",&a,&b,&c);//接收数据;
            cap[a][b]+=c;
            //可能存在多条重复路径;
        }
        karp();
    }
    return 0;
}