HDU 3549 基础网络流EK算法 Flow Problem

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Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 10184    Accepted Submission(s): 4798

Problem Description

Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

 

Input

The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)

 

Output

For each test cases, you should output the maximum flow from source 1 to sink N.

 

Sample Input

2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1

 

Sample Output

Case 1: 1 Case 2: 2

 

Author

HyperHexagon

 

Source

HyperHexagon's Summer Gift (Original tasks)

 

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#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int dp[100][100],pre[100];
const int tmin=999999999;
int maxflow;
void EK(int start,int end,int n){
    while(1){
        queue<int>q;
        q.push(1);
       int  minflow=tmin;
        memset(pre,0,sizeof(pre));
        while(!q.empty()){
            int u=q.front();
            q.pop();
            for(int i=1;i<=n;i++){
                if(dp[u][i]>0&&!pre[i]){
                    pre[i]=u;
                    q.push(i);
                }
            }
        }
        if(pre[end]==0)
            break;
        for(int i=end;i!=start;i=pre[i]){
            minflow=min(dp[pre[i]][i],minflow);
        }
        for(int i=end;i!=start;i=pre[i]){
            dp[pre[i]][i]-=minflow;
            dp[i][pre[i]]+=minflow;
        }
        maxflow+=minflow;
    }
}
int main(){
   int count=0;
   int n,m;
   int t;
   scanf("%d",&t);
   while(t--){
        scanf("%d%d",&n,&m);
       memset(dp,0,sizeof(dp));
       memset(pre,0,sizeof(pre));
       count++;
       int u,v,w;
       for(int i=1;i<=m;i++){
           scanf("%d%d%d",&u,&v,&w);
           dp[u][v]+=w;
       }
       maxflow=0;
       EK(1,n,n);
       printf("Case %d: %d
",count,maxflow);
   }
   return 0;
}