hdu 4324 Triangle LOVE(拓扑排序)

Triangle LOVE

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 16   Accepted Submission(s) : 11

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Problem Description

Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world!
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
  Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.

Input

The first line contains a single integer t (1 <= t <= 15), the number of test cases.
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). A_i,j = 1 means i-th people loves j-th people, otherwise A_i,j = 0.
It is guaranteed that the given relationship is a tournament, that is, A_i,i= 0, A_i,j ≠ A_j,i(1<=i, j<=n,i≠j).

Output

For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”.
Take the sample output for more details.

Sample Input

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

Sample Output

Case #1: Yes
Case #2: No

#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;
int i,j,n,t;
int cnt[2002];
char mp[2002][2002];
bool toposort()
{
    queue<int> Q;
    int num=0;
    for(int i=0;i<n;i++) if (cnt[i]==0) Q.push(i);
    if(Q.empty()) return 0;
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        num++;
        for(int i=0;i<n;i++)
        if (mp[u][i]=='1')
        {
            cnt[i]--;
            if (cnt[i]==0) Q.push(i);
            if (cnt[i]<0) return 0;
        }
    }
    if (num<n) return 0;
    return 1;
}
int main()
{
    scanf("%d",&t);
    for(int ii=1;ii<=t;ii++)
    {
        scanf("%d",&n);
        memset(cnt,0,sizeof(cnt));
        for(i=0;i<n;i++)
        {
            scanf("%s",mp[i]);
            for(j=0;j<n;j++)
                if (mp[i][j]=='1') cnt[j]++;
        }
        printf("Case #%d: ",ii);
        if (!toposort()) printf("Yes\n");
                 else printf("No\n");
    }
    return 0;
}