HDOJ-4324-Triangle LOVE（拓扑排序）

最新推荐文章于 2018-11-07 21:43:09 发布

原创最新推荐文章于 2018-11-07 21:43:09 发布 · 468 阅读

0 ·

CC 4.0 BY-SA版权

拓扑排序专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一个用于检测人群中是否存在三角恋关系的算法，通过输入人际关系矩阵，输出是否存在三角恋。

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): 3668 Accepted Submission(s): 1439

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

判断有没有三角恋（其实多角恋也算）

拿队列排序，借排序看有多少能排序的，出现环时不能排，无环时:总人数-排序计数==0

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<functional>
#define max 2100
using namespace std;
int map[max][max],cnt;
int indegree[max]; 
void topo(int n)
{
	priority_queue<int,vector<int>,greater<int> >Q;
    int i,j,m,t=0;
    for(i=1;i<=n;++i){
        if(indegree[i]==0){
             Q.push(i);
        }
    }
    while(!Q.empty()){
    	int top=Q.top();Q.pop();
    	indegree[top]=-1;
    	cnt++;//排一个加一个
 		for(i=1;i<=n;++i){
    		if(map[top][i]){
    			indegree[i]--;
    			if(indegree[i]==0)
    				Q.push(i);
			}
		}
    }
}
int main()
{
    int n,T,t;
    char x[max];
    scanf("%d",&T);
    for(t=1;t<=T;t++){
    	scanf("%d",&n);
        memset(indegree,0,sizeof(indegree));
        memset(map,0,sizeof(map));
		for(int i=1;i<=n;i++){
			scanf("%s",x);
			for(int j=1;j<=n;j++){
				map[i][j]=x[j-1]-'0';
				if(map[i][j])
					indegree[j]++;
			}
		}
		cnt=0;
        topo(n);
        if(n-cnt)
        	printf("Case #%d: Yes\n",t);
		else
			printf("Case #%d: No\n",t);
    }
    return 0;
}

顺便写了个dfs版,结果超时了,一起贴出来

//dfs   TLE
/*
#include<cstdio>
#define max 2100
int map[max][max],N,deep,star,loop,t;
void dfs(int i){
	if(deep>3)
		return;
	if(deep==3&&map[i][star]){
		loop=1;
		return;
	}
	for(int j=1;j<=N;j++){
		if(map[i][j]){
			deep++;
			dfs(j);
			deep--;
		}
	}
}
int main(){
	int T;
	char x[max];
	scanf("%d",&T);
	for(t=1;t<=T;t++){
		scanf("%d",&N);
		for(int i=1;i<=N;i++){
			scanf("%s",x);
			for(int j=1;j<=N;j++){
				map[i][j]=x[j-1]-'0';
			}
		}
		deep=1;
		loop=0;
		for(int i=1;i<=N;i++){
			dfs(i);
			star=i;
		}
		if(loop)
			printf("Case #%d: Yes\n",t);
		else
			printf("Case #%d: No\n",t);
	}
	return 0;
}
*/