hdu3549 Flow Problem（裸最大流）

 

                                              Flow Problem

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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 

题目比较简单，入门级最大流问题。

#include<cstdio>
#include<algorithm>
#include<queue>
#include<string.h>
#define MAXN 21
#define INF 1<<28

using namespace std;

int cap[MAXN][MAXN], pre[MAXN], vis[MAXN], maxflow[MAXN];
int Maxflow;
int n, m, i, j;

void init()
{
    int x, y, z;
    memset(cap, 0, sizeof(cap));
    memset(pre, 0, sizeof(pre));

    scanf("%d%d", &n, &m);
    for(i=1; i<=m; i++)
    {
        scanf("%d%d%d", &x, &y, &z);
        cap[x][y] += z;//重边
    }
}

int min(int a, int b)
{
    return a>b?b:a;
}

int main()
{
    int T, u, k=1;
    scanf("%d", &T);
    while(T--)
    {

        init();
        queue<int>q;
        Maxflow = 0;

        while(1)
       {
        memset(maxflow, 0, sizeof(maxflow));
        memset(vis, 0, sizeof(vis));
        maxflow[1] = INF;


        q.push(1);
        while(!q.empty())
        {
             u = q.front();
             q.pop();
             for(int v=1; v<=n; v++)
             {
                 if(!vis[v] && cap[u][v]>0)
                    {
                        vis[v] = 1;
                        pre[v] = u;
                        q.push(v);
                        maxflow[v] = maxflow[u]<cap[u][v]?maxflow[u]:cap[u][v];
                    }
             }
        }
        if(maxflow[n]==0) break;

        for(int i=n; i!=1; i=pre[i])
        {
            cap[pre[i]][i]-=maxflow[n];
            cap[i][pre[i]]+=maxflow[n];
        }

       Maxflow += maxflow[n];
    }
      printf("Case %d: %d\n", k++, Maxflow);
    }
     return 0;
}