Triangle LOVETime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3508Accepted Submission(s): 1364
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
25001001000001001111011100050111100000010000110001110
Sample Output
Case #1: YesCase #2: No
Author
判断三角恋，第一行的意思就是0喜欢1,2,3,4；
第二行是1喜欢0；
第三行就是3喜欢1和4，，依次类推：
附ac代码：
#include<stdio.h>#include<string.h>int topo[2020],i,n,m,j,k,l,t,map[2020][2020];char c[2020][2020];int main(){int biaoji=1;scanf("%d",&t);while(t–){scanf("%d",&n);memset(topo,0,sizeof(topo));for(i=0;i<n;i++)scanf("%s",&c[i]);for(i=0;i<n;i++)for(j=0;j<n;j++){if(c[i][j]=='1')topo[j]++;}int flag=0;for(i=0;i<n;i++){if(topo[i]==0){topo[i]=1000;//经调试，如果i有三角恋关系类似的for(j=0;j<n;j++)//则topo[i]一定减不到0if(c[i][j]!='0')topo[j]–;flag++;//则flag+不到ni=0;}}printf("Case #%d: ",biaoji++);if(flag==n)printf("No\n");elseprintf("Yes\n");}return 0;}
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