HDU3549_Flow Problem(网络流/EK)

Posted 2020-09-30 liguangsunls

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6987    Accepted Submission(s): 3262

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

解题报告

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 30
#define inf 99999999
using namespace std;
int n,m,a[N],flow,pre[N];
int Edge[N][N];
queue<int >Q;
int bfs()
{
    while(!Q.empty())
    Q.pop();
    memset(a,0,sizeof(a));
    memset(pre,0,sizeof(pre));
    Q.push(1);
    pre[1]=1;
    a[1]=inf;
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        for(int v=1;v<=n;v++)
        {
            if(!a[v]&&Edge[u][v]>0)
            {
                pre[v]=u;
                a[v]=min(Edge[u][v],a[u]);
                Q.push(v);
            }
        }
        if(a[n])break;
    }
    if(!a[n])
    return -1;
    else return a[n];
}
void ek()
{
    int a,i;
    while((a=bfs())!=-1)
    {
        for(i=n;i!=1;i=pre[i])
        {
            Edge[pre[i]][i]-=a;
            Edge[i][pre[i]]+=a;
        }
        flow+=a;
    }
}
int main()
{
    int t,i,j,u,v,w,k=1;
    scanf("%d",&t);
    while(t--)
    {
        flow=0;
        memset(Edge,0,sizeof(Edge));
        scanf("%d%d",&n,&m);
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            Edge[u][v]+=w;
        }
        ek();
        printf("Case %d: ",k++);
        printf("%d\n",flow);
    }
    return 0;
}