HDU 4324：Triangle LOVE【拓扑排序】

最新推荐文章于 2019-02-15 19:53:27 发布

林次次

最新推荐文章于 2019-02-15 19:53:27 发布

阅读量840

点赞数

CC 4.0 BY-SA版权

分类专栏：＝＝图论＝＝＝＝＝＝C&＆C++＝＝＝＝拓扑排序文章标签：图论拓扑排序 HDU

本文链接：https://blog.youkuaiyun.com/lin14543/article/details/47806743

＝＝＝＝C&＆C++＝＝＝＝同时被 3 个专栏收录

153 篇文章

订阅专栏

＝＝图论＝＝

25 篇文章

订阅专栏

拓扑排序

4 篇文章

订阅专栏

该博客介绍了HDU 4324题目——Triangle LOVE的解决方法，主要涉及图论中的拓扑排序概念。文章首先强调理解题意，指出判断是否存在环是关键，存在环则输出Yes，否则输出No。博主分享了通过拓扑排序实现的AC代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3692 Accepted Submission(s): 1455

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

首先要理解题意～～～判断是否有成环的～～若有则输出Yes，木有则输出No~~

AC-code:

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
int n,map[2010][2010],ind[2010];
char str[2010];
void topo()
{
	int i,top,k=0;
	queue<int>q;
	for(i=0;i<n;i++)
		if(!ind[i])
			q.push(i);
	while(!q.empty())
	{
		top=q.front();
		q.pop();
		ind[top]=-1;
		k++;
		for(i=0;i<n;i++)
		{
			if(map[top][i])
				ind[i]--;
			if(ind[i]==0)
				q.push(i);
		}	
	}
	if(k==n)
		printf("No\n");
	else
		printf("Yes\n");
}
int main()
{
	int t,k,i,j;
	scanf("%d",&t);
	for(k=1;k<=t;k++)
	{
		memset(map,0,sizeof(map));
		memset(ind,0,sizeof(ind));
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%s",str);
			for(j=0;j<n;j++)
			{
				map[i][j]=str[j]-'0';
				if(map[i][j])
					ind[j]++;
			}
		}
		printf("Case #%d: ",k);
		topo();
	}
	return 0;
}