joj 1169

这道题着实让我蛋疼了一下，看完书上的代码之后信心满满还稍微修改了一下方法来做这道题，结果wa的暗无天日，后来猛然想到oj的测试数据不一定一半是零，然后感觉样例输入果然很坑人，然后AC：

#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
int cap[32][32];
int flow[32][32];
int a[32];
int p[32];
int main()
{
    int n;
    while(scanf("%d",&n),n)
    {
        memset(flow,0,sizeof(flow));
        memset(cap,0,sizeof(cap));
        int i,j;
        int f=0;
        for(i=1;i<=n;i++)
           for(j=1;j<=n;j++)
           {
               int m;
               scanf("%d",&m);
               cap[i][j]+=m;
               cap[j][i]+=m;
               flow[j][i]+=m;
           }
        if(n==1)
            printf("Maximum number of goods: 0.\n");
        int flag=0;
        while(1)
        {
            memset(a,0,sizeof(a));
            a[1]=2147483647;
            queue<int> q;
            q.push(1);
            while(!q.empty())
            {
                int k=q.front();
                q.pop();
                for(i=1;i<=n;i++)
                    if(a[i]==0&&cap[k][i]>flow[k][i])
                    {
                        p[i]=k;
                        q.push(i);
                        a[i]=((a[k]<(cap[k][i]-flow[k][i]))?a[k]:(cap[k][i]-flow[k][i]));
                    }
            }
            if(a[n]==0)
                 break;
            for(i=n;i!=1;i=p[i])
            {
                flow[p[i]][i]+=a[n];
                flow[i][p[i]]-=a[n];
            }
            f+=a[n];
        }
        printf("Maximum number of goods: %d.\n",f);
    }
}