HDOJ-----4324Triangle LOVE---拓扑排序

最新推荐文章于 2018-11-07 21:43:09 发布

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本文介绍了一道关于“TriangleLove”的编程竞赛题，探讨了如何通过拓扑排序判断是否存在三角形爱情关系。利用邻接矩阵表示人与人之间的关系，并采用队列实现拓扑排序，最终确定是否存在环。

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Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4700 Accepted Submission(s): 1852

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

最开始以为是二分图判断，后来才知道是拓扑判断环

一群人，一个人必定有喜欢的人，可以看做是排序，单人不会成环，题目规定两人不会互相喜欢，双人也不会成环

三环及以上环必定符合题意，即判断是否存在环即可

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CL(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int maxn = 2e3+10;
const int MOD = 1e9+7;
vector<int > edge[maxn];  
int in[maxn];  
int n, ans;  
char s[maxn];
void topo(){  
    queue<int > Q;  
    for(int i = 0; i < n; i++){  
        if(!in[i]) Q.push(i);  
    } 
    while(!Q.empty()){  
        int u = Q.front();  
        Q.pop();  
        in[u] = -1; 
        ans++;  //ans记录人数，出现环就会少人，则ans != n
        for(int i = 0; i < edge[u].size(); i++){  
            int v = edge[u][i];  
            in[v]--;  
            if(!in[v]) Q.push(v);  
        }  
    }  
}  
int main(){  
    int t, kcase = 1;
    scanf("%d", &t);  
    while(t--){  
        scanf("%d", &n); 
        for(int i = 0; i <= n; i++) {
        	in[i] = 0;
        	edge[i].clear(); 
		} 
        for(int i = 0; i < n; i++){  
        	scanf(" %s", s);
	        for(int j = 0; j < n; j++){
	        	//scanf(" %c", &ch); //用%c输入就会超时
	            if(s[j] == 49){
		            edge[i].push_back(j);  
		            in[j]++;  
				}
			}
        }  
        ans = 0;  
        topo();  
        printf("Case #%d: %s\n", kcase++, ans != n ? "Yes" : "No");  
    }  
    return 0;  
}