hdu 4324 Triangle LOVE

最新推荐文章于 2019-02-15 19:53:27 发布

threeh20

最新推荐文章于 2019-02-15 19:53:27 发布

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分类专栏：拓扑排序

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本文介绍了一道关于图论的算法题TriangleLOVE，题目要求判断N个人中是否存在三角恋关系。通过输入的人际关系矩阵，利用拓扑排序的方法来解决此问题。详细解释了拓扑排序的过程及其实现。

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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

Sample Output

  
  
   
   Case #1: Yes
Case #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;
 }