poj 1273

Drainage Ditches

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 12   Accepted Submission(s) : 7

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

   
   
   
   

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

   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    50
   
   
   
   

求源点1到汇点m的最大流  ek算法

第一次最大流。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        int c[205][205],f[205][205];
        int vis[205],p[205];
        queue<int >q;
        memset(c,0,sizeof(c));  //c标记容量
        memset(f,0,sizeof(f));  //f标记流量
        for(int i=0;i<n;i++)
        {
            int aa,bb,cc;
            scanf("%d%d%d",&aa,&bb,&cc);
            if(aa!=bb)  
            c[aa][bb]+=cc;  //可能会有重复边
        }
        int ff=0;
        while(1)
        {
            memset(vis,0,sizeof(vis));  //记录当前点的流量
            memset(p,-1,sizeof(p));     //记录当前点的前驱
            vis[1]=INF;
            q.push(1);
            while(q.size())
            {
                int st=q.front();
                q.pop();
                for(int i=1;i<=m;i++)
                {
                    if(!vis[i]&&f[st][i]<c[st][i])
                    {
                        vis[i]=min(vis[st],c[st][i]-f[st][i]);      //流量为当前路径的最大流量或者为当前点的最大流量
                        p[i]=st;                                    //记录前驱
                        q.push(i);
                    }
                }
            }

            if(!vis[m]) break;

            for(int i=m;i!=1;i=p[i])  
            {
                f[p[i]][i]+=vis[m];
                f[i][p[i]]-=vis[m];
            }
            ff+=vis[m];
        }

        printf("%d\n",ff);
    }
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long

int main()
{
    int n,m;
    int t,tt=1;
    int c[205][205],f[205];
    bool vis[205];
    int p[205];
    queue<int >q;
    //scanf("%d",&t);
    while(~scanf("%d%d",&n,&m))
    {
        memset(c,0,sizeof(c));  //c标记容量
        for(int i=0; i<n; i++)
        {
            int aa,bb,cc;
            scanf("%d%d%d",&aa,&bb,&cc);
            //if(aa!=bb)
                c[aa][bb]+=cc;  //可能会有重复边
        }
        int ff=0;
        while(1)
        {
            memset(vis,false ,sizeof(vis));  //记录当前点的流量
            memset(p,-1,sizeof(p));          //记录当前点的前驱
            for(int i=1;i<=m;i++)
                f[i]=INF;
            vis[1]=true ;
            while(q.size()) q.pop();
            q.push(1);
            while(q.size())
            {
                int st=q.front();
                q.pop();
                if(st==m) break;
                for(int i=1; i<=m; i++)
                {
                    if(!vis[i]&&c[st][i]>0)
                    {
                        vis[i]=true ;
                        f[i]=min(f[st],c[st][i]);                   //流量为当前路径的最大流量或者为当前点的最大流量
                        p[i]=st;                                    //记录前驱
                        q.push(i);
                    }
                }
            }

            if(!vis[m]) break;
            //cout<<f[n]<<endl;

            int st=m;
            while(st!=1)
            {
                int ed=p[st];
                c[ed][st]-=f[m];    //正向边
                c[st][ed]+=f[m];    //反向边
                st=ed;
            }

            ff+=f[m];
        }

        printf("%d\n",ff);
    }
}