Flow Problem

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

Case 1: 1
Case 2: 2

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;
}