Triangle LOVE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2663 Accepted Submission(s): 1070
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”.
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).
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.
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
/*
题解:dfs解决,参考网上的代码,一开始以为仅仅需要判断是否有环果断wrong
后来发现题意说“Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
参考网上的dfs解决,但仍然不是很理解。
*/
#include<cstdio> #include<cstring> #define maxn 2002 int vis[maxn],dis[maxn],flag,n; char map[maxn][maxn]; void dfs(int step) { if(flag) return; vis[step]=1; for(int i=1; i<=n; i++) { if(map[step][i]=='1') { if(vis[i]&&dis[step]==dis[i]+2) { flag=1; break; } else if(!vis[i]) { dis[i]=dis[step]+1; dfs(i); } } } } int main() { int T,cnt=1; scanf("%d",&T); while(T--) { // memset(vis,0,sizeof(0));事前初始化就会wrong scanf("%d",&n); //getchar(); for(int i=1; i<=n; i++)//字符串这样可以从map[1][1]开始存储 { vis[i]=0; //必须要这样初始化 scanf("%s",map[i]+1);//从1开始存储 } dis[1]=0; flag=0; dfs(1);//从1开始搜索 printf("Case #%d: ",cnt++); if(flag) printf("Yes\n"); else printf("No\n"); } return 0; }