【杭电4324】 Triangle LOVE

B - Triangle LOVE

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

Input

Output

Sample Input

Sample Output

Hint

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

25001001000001001111011100050111100000010000110001110

Sample Output

Case #1: YesCase #2: No

题意分析(转载)：
此题可以一遍拓扑排序判环求解 即只需要找到一个环，
就必定存在三元环 证明如下： 假设存在一个n元环，
因为a->b有边，b->a必定没边，反之也成立
所以假设有环上三个相邻的点a-> b-> c,那么如果c->a间有边，
就已经形成了一个三元环，如果c->a没边，那么a->c肯定有边，
这样就形成了一个n-1元环。。。。
所以只需证明n大于3时一定有三元环即可，显然成立。
*/

方法：拓扑排序

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int in[2000];char s[2000][2000];int main(){int t,n;scanf("%d",&t);int ch=1;while(t--){memset(in,0,sizeof(in));int faut=0;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%s",&s[i]);for(int k=0;k<n;k++)if(s[i][k]=='1')in[k]++;}for(int i=0;i<n;i++){int k;for(k=0;k<n;k++)if(in[k]==0) break;if(k==n) {faut=1;break;}else{in[k]--;for(int g=0;g<n;g++)if(s[k][g]=='1'&&in[g]!=0)in[g]--;}}if(faut==1)printf("Case #%d: Yes\n",ch++); else printf("Case #%d: No\n",ch++); } return 0; }