hdu 4324 Triangle LOVE

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4875    Accepted Submission(s): 1923

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

25001001000001001111011100050111100000010000110001110

 

Sample Output

Case #1: YesCase #2: No

 

就像题目名字一样，给出一张图，问有没有三角恋关系，也就是有无成环。。

#include<iostream>#include<cstring>#include<cmath>#include<algorithm>#include<cstdio> using namespace std;char ma1[2221][2221];int inde[2222];int flag;int ans[11111];int n,m;bool tapu(){    int i,j;    for(i=0;i<n;i++)    for(j=0;j<n;j++)    {        if(inde[j]==0)        {            inde[j]=-1;            ans[i]=j;            for(int s=0;s<n;s++)            {                if(ma1[j][s]=='1')                inde[s]--;            }            break;        }        if(j==n-1)        {            return 0;        }    }    return 1;}int main(){    int T;    cin>>T;    int case1=0;    while(T--)    {        case1++;        cin>>n;        int i,j,k;        flag=1;int a,b;        memset(ma1,0,sizeof(ma1));        memset(inde,0,sizeof(inde));        for(i=0;i<n;i++)        {                    scanf("%s",&ma1[i]);        for(int s=0;s<n;s++)        if(ma1[i][s]=='1')        inde[s]++;                }            cout<<"Case #"<<case1<<": ";         if(tapu())         {             cout<<"No"<<endl;         }         else cout<<"Yes"<<endl;    }    return 0; }