Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2455    Accepted Submission(s): 997

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

不知道为什么只要成环就包括三角环。。

#include <stdio.h>
#include <string.h>
#define maxn 2002

bool map[maxn][maxn];
char buf[maxn];
int indegree[maxn], queue[maxn];

void addEdge(int n)
{
	int i, j;
	for(i = 0; i < n; ++i){
		scanf("%s", buf);
		for(j = 0; j < n; ++j)
			if(buf[j] == '0') map[i][j] = 0;
			else{
				map[i][j] = 1;
				++indegree[j];
			}
	}
}

bool topoSort(int n)
{
	int i, u, front = 0, back = 0;
	for(i = 0; i < n; ++i)
		if(!indegree[i]) queue[back++] = i;
	while(front != back){
		u = queue[front++];
		for(i = 0; i < n; ++i){
			if(map[u][i] && !--indegree[i])
				queue[back++] = i;
		}
	}
	return back == n;
}

int main()
{
	int t, n, cas = 1;
	scanf("%d", &t);
	while(t--){
		memset(indegree, 0, sizeof(indegree));
		scanf("%d", &n); addEdge(n);
		printf("Case #%d: ", cas++);
		if(!topoSort(n)) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

HDU4324 Triangle LOVE 【拓扑排序】