hdoj 4324 Triangle LOVE 【拓扑】

Triangle LOVE

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

Total Submission(s): 2586    Accepted Submission(s): 1051

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

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

 

Sample Output

Case #1: Yes
Case #2: No

题意：给出你全部人的关系。然后让你推断一下是否存在三角恋或多角恋。

分析：我们能够依据关系建一个有向图，假如A喜欢B那么就让A指向B。假如存在三角恋或多角恋那么肯定会形成一个环，我选择用拓扑排序，假设形成环的话肯定不会把全部的数都排序。

注意：用链式前向星的话，边的数组要开到2000*1999之上。。RE了好几次。

心得: 比赛的时候没有想到拓扑。仅仅是想到了并查集。左后还是没有做出来。

并查集能够推断一棵树是不是成环，拓扑才干够推断一个图是不是成环。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 2005
struct node{
	int to, next;
}s[20000005];
int in[M], head[M], n, tot, queue[M];
char map[M][M];
void getmap(int a, int b){
	s[tot].to = b;
	in[b]++;
	s[tot].next = head[a];
	head[a] = tot++;
}
int toposort(){
	bool vis[M];
	memset(vis, false, sizeof(vis));
	int i, j, iq = 0;
	for(i = 0; i < n; i ++){
		if(!in[i]){
			queue[iq++] = i;
			vis[i] = 1;
		}
	}
	for(i = 0; i < iq; i++){
		int temp = head[queue[i]];
		for(j = temp; j != -1; j = s[j].next){
			 if(!(--in[s[j].to])){
				queue[iq++] = s[j].to;
			}
		}
	}
	if(iq < n) return 1;
	else return 0;
}
int main(){
	int t, v = 1;
	scanf("%d", &t);
	while(t --){
		memset(in, 0, sizeof(in));
		memset(head, -1, sizeof(head));
		tot = 0;
		scanf("%d", &n);
		int i, j, flag;
		for(i = 0; i < n; i ++){
			scanf("%s", map[i]);
			for(j = 0; j < n; j ++){
				if(map[i][j] == '1'){
					getmap(i, j);
				}
			}
		}
		printf("Case #%d: ", v++);
		flag = toposort();
		if(flag)  printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

版权声明：本文博客原创文章。博客，未经同意，不得转载。