hdu 4324

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2478    Accepted Submission(s): 1011

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

 对于两个任何两个点都会有一条边将他们相连（A_i,j ≠ A_j,i）

假设目前有一个环大于3，其中有相连的三个点A,B,C，即 A->B,B->C

如果C->A 有一条边，就形成了一个3环

如果C->A 没有边，那么一定有 A->C，从而形成一个n-1环，以这种方式一定可以形成一个3环

所以只要判断这些点构成一个环即可

#include<string>
#include<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
#include<cstring>
#include<stdlib.h>
#include<string>
#include<cmath>
using namespace std;
#define pb push_back
vector<int >p[10010];
int in[10010],n,m,deep[10010],num[10010];
char dp[2010][2010];
void init(){
    memset(in,0,sizeof(in));
    memset(deep,0,sizeof(deep));
    memset(num,0,sizeof(num));
    for(int i=0;i<=n;i++) p[i].clear();
}
int tuopu(){
    queue<int >ak_47;
    int cnt=0;
    for(int i=1;i<n;i++)
    if(in[i]==0) ak_47.push(i),deep[i]=1,num[1]++;
    while(!ak_47.empty()){
        int pos=ak_47.front();
        cnt++;
        for(int i=0;i<p[pos].size();i++){
            int to=p[pos][i];
            if(--in[to]==0) ak_47.push(to),deep[to]=deep[pos]+1,num[deep[to] ]++;
        }
        ak_47.pop();
    }
    if(cnt<n) return 1;
    else return 0;

}
int main(){
    int cas=0,t;
    cin>>t;
    while(t--){
        cin>>n;
        init();
        for(int i=1;i<=n;i++){
            scanf("%s",dp[i]);
            for(int j=0;j<n;j++)
            if(dp[i][j]=='1') in[j+1]++,p[i].pb(j+1);
        }
        if(tuopu()) printf("Case #%d: Yes\n",++cas);
        else   printf("Case #%d: No\n",++cas);
    }
}