【HDU】4324 - Triangle LOVE（拓扑）_hdu triangle love-优快云博客

本文介绍了一种判断在特定条件下是否存在‘Triangle Love’关系的算法。通过构建拓扑排序来解决给定N个人之间的关系矩阵中是否存在三角恋的问题。

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4672 Accepted Submission(s): 1839

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

Sample Output

  
   Case #1: Yes
Case #2: No

Author

BJTU

Source

2012 Multi-University Training Contest 3

由题目已给的条件已知，如果四个以上的点构成回路，那么肯定有三角恋关系。那么拓扑就行了。

注意输入要用%s，否则会超时。

代码如下：

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
#define MAX 2000
vector<int> d[MAX+11];
int in[MAX+5];
bool ans;
int n;
void init()
{
	for (int i = 1 ; i <= n ; i++)
		d[i].clear() , in[i] = 0;
}
void topo()
{
	queue<int> q;
	for (int i = 1 ; i <= n ; i++)
		if (in[i] == 0)
			q.push(i);
	while (!q.empty())
	{
		int pos = q.front();
		q.pop();
		in[pos] = -1;
		for (int i = 0 ; i < d[pos].size() ; i++)
		{
			int dot = d[pos][i];
			in[dot]--;
			if (!in[dot])
				q.push(dot);
		}
	}
}
int main()
{
	int u;
	char str[2016];
	int Case = 1;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d",&n);
		init();
//		getchar();
		for (int i = 1 ; i <= n ; i++)
		{
			scanf ("%s",str+1);
			for (int j = 1 ; j <= n ; j++)
			{
//				int t;
//				scanf ("%c",&t);		//这么输入超时 
				if (str[j] == '1')
					d[i].push_back(j) , in[j]++;
			}
//			getchar();
		}
		topo();
		ans = false;
		for (int i = 1 ; i <= n ; i++)
		{
			if (in[i] > 0)
			{
				ans = true;
				break;
			}
		}
		printf ("Case #%d: %s\n",Case++,ans ? "Yes" : "No");
	}
	return 0;
}