hdu4324(拓扑排序)

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2138    Accepted Submission(s): 898

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
   
   
   
   

 

Author

BJTU

题解：

      本题要求给定的有向图里是否又a->b->c->a的三角形环。由于矩阵中任意两点有么有正向边，要么有反向边，所以只要有环，就能构成三角形环。证明如下：假设有n元环，a->b->c……->a，若c->a有边，则存在三角形环a->b->c->a；若无边，则说明不存在c->a的边从而存在a->c的边，于是有n元环a->b->c……->a变成有n-1元环，a->c……->a，最终可以证明存在三角形环。时间复杂度为拓扑排序的O(n*n)，在本题可以接受。

#include<iostream>
#include<cstring>
using namespace std;

const int MAXN=2000+10;
int inDev[MAXN];
bool visited[MAXN];
bool g[MAXN][MAXN];
char str[MAXN];
int n;

void init()
{
	memset(g,0,sizeof(g));
	memset(visited,0,sizeof(visited));
	memset(inDev,0,sizeof(inDev));
}

void input()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%s",str);
		for(int j=0;j<n;j++)
		{
			if('1'==str[j])
			{
			g[i][j]=true;
			++inDev[j];
			}
		}
	}
}

void topSort()  
{  
    int i,tag=0,k=0;  
    while(k<n)  
    {  
		for(i=0;i<n;i++)
			if(!visited[i]&&0==inDev[i])
				break;
		if(i>=n)
		{
			tag=1;
			break;
		}
		++k;
		visited[i]=true;
		for(int j=0;j<n;j++)
			if(g[i][j]&&inDev[j]!=0)
				--inDev[j];
    }  
	if(tag)
		printf("Yes\n");
	else printf("No\n");
}  

int main()
{
	int cas,tag=0;
	cin>>cas;
	while(cas--)
	{
		init();
		input();
		printf("Case #%d: ",++tag);
		topSort();
	}
	system("pause");
	return 0;
}