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hdoj 4324 Triangle LOVE 【拓扑】

2024-07-26 21:26:17   219人阅读

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). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,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：给出你所有人的关系，然后让你判断一下是否存在三角恋或多角恋。

分析：我们可以根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，如果形成环的话肯定不会把所有的数都排序。

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，只是想到了并查集，左后还是没有做出来。

并查集可以判断一棵树是不是成环，拓扑才可以判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;
}


hdoj 4324 Triangle LOVE 【拓扑】


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