HDU 3549 Flow Problem

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2857    Accepted Submission(s): 1348

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)

 //有一网络流入门模板题目，给下届的小朋友又一敲代码练习的机会

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zhengfeng

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#define N 1003
using namespace std;
int flow[N][N],cap[N][N];
int f[N];
int n,m;
int BFS()
{
    int u,v;
    int a[N];
    int sc=0;
    for(int i=1;i<=n;i++)//把里面的这种初始化过程改成memset，那么将变成4900+Ms，而这样就60+Ms
          for(int j=1;j<=n;j++)//相差好大呀
             flow[i][j]=0;
    for(;;)
    {
        queue<int>Q;
        for(int i=1;i<=n;i++)
           a[i]=0;
        a[1]=N;
        Q.push(1);
        while(!Q.empty())
        {
            u=Q.front();Q.pop();
             for(v=1;v<=n;v++)
               if(!a[v]&&cap[u][v]-flow[u][v]>0)
              {
                  f[v]=u;
                  Q.push(v);
                  a[v]=a[u]<cap[u][v]-flow[u][v]?a[u]:cap[u][v]-flow[u][v];
              }

        }
        if(a[n]==0) break;
        for(u=n;u!=1;u=f[u])
        {
            flow[f[u]][u]+=a[n];
            flow[u][f[u]]-=a[n];
        }
        sc+=a[n];
    }
    return sc;
}
int main()
{
    int t,T=1;
    int u,v,c;
    scanf("%d",&t);
    while(t--)
    {
       scanf("%d%d",&n,&m);
       for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
             cap[i][j]=0;
       while(m--)
       {
           scanf("%d%d%d",&u,&v,&c);
           cap[u][v]+=c;
       }
       printf("Case %d: %d\n",T++,BFS());
    }
    return 0;
}