HDU 4323 Triangle LOVE【拓扑排序】

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本文介绍了一种判断在特定人际关系网络中是否存在三角恋关系的算法。该算法通过遍历关系矩阵并利用拓扑排序原理来确定是否存在A爱B、B爱C且C爱A的三角恋模式。

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Triangle LOVE

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

Total Submission(s): 4620 Accepted Submission(s): 1816

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

AC代码：

#include<cstdio>
#include<cstring>
#include<stack>
#include<algorithm>

using namespace std;

const int MAXN=2e3+12;
int in[MAXN]; 
char G[MAXN][MAXN];

bool Topo(int N)
{
    int cnt=0;
    stack<int> S;
    for(int i=1;i<=N;++i) {
    	if(!in[i]) S.push(i);
	} 
	while(!S.empty() ) {
		int tem=S.top(); S.pop();
		++cnt;
		//.printf("ddd\n");
		for(int i=1;i<=N;++i) {
			if(G[tem][i]=='1') {
				--in[i];
			    if(!in[i]) S.push(i);//度数原本为0的 不能入队 
			}
		}
	}
	return cnt!=N?true:false;
}
int main()
{
	int N,T,Kase=0;
	scanf("%d",&T);
	while(T--) {
		scanf("%d",&N);
		memset(in,0,sizeof(in));
		for(int i=1;i<=N;++i) {
			scanf("%s",G[i]+1);
			for(int j=1;j<=N;++j) {
				if(G[i][j]=='1') ++in[j];
 			} 
		} 
		printf("Case #%d: ",++Kase);
		printf("%s\n",Topo(N)?"Yes":"No");
	}
	return 0;
}