hdoj 4324 Triangle LOVE【拓扑排序判断是否存在环】

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3566    Accepted Submission(s): 1395

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

 

题意：给出一个矩阵s[][]如果其中s[i][j]等于1代表i喜欢j，（但是题目保证不可能有两个人互相相爱）现在问这些人的关系中是否存在三角           恋（A爱B，B爱C，C爱A）如果有则输出yes否则输出no

题解：因为题目已经保证不会有两个人互相相爱，所以说成环最少也是三人环，直接拓扑排序判断是否成环即可（要用邻接表做，矩阵会超           时）

#include<stdio.h>
#include<string.h>
#include<vector>
#include<queue>
#define MAX 2010
using namespace std;
int n,k;
vector<int>map[MAX];
char s[MAX];
int vis[MAX];
void getmap()
{
	int i,j;
	for(i=1;i<=n;i++)
	    map[i].clear();
	for(i=1;i<=n;i++)
	{
		scanf("%s",s);
	    for(j=0;j<n;j++)
	    {
	    	if(s[j]=='1')
	    	{
	    		map[i].push_back(j+1);
	    		    vis[j+1]++;
			}
		}
	}
}

void tuopu()
{
	int i,j;
	queue<int>q;
	while(!q.empty())
	    q.pop();
	for(i=1;i<=n;i++)
	    if(vis[i]==0)
	        q.push(i);
	int u,v;
	int ans=0;
	while(!q.empty())
	{
		u=q.front();
		ans++;
		q.pop();
		for(i=0;i<map[u].size();i++)
		{
		    v=map[u][i];
		    vis[v]--;
		    if(vis[v]==0)
		    q.push(v);
		}
	}
	printf("Case #%d: ",k++);
	if(ans!=n)
	printf("Yes
");
	else
	printf("No
");
}
int main()
{
	int t,i,j;
	k=1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		memset(vis,0,sizeof(vis));
	    getmap();
	    tuopu();
	}
	return 0;
}