HDU 4324 Triangle LOVE (拓扑排序)

Triangle LOVE

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 25   Accepted Submission(s) : 15

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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

AC代码

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int n;
char map[2001][2001];
int indegree[2001];
int cnt;
void toposort()
{
    for(int i=1;i<=n;i++)
    {
        int flag=0;
        for(int j=1;j<=n;j++)
        {
            if(indegree[j]==0)
            {
                cnt++;
                flag=1;
                indegree[j]--;
                for(int k=1;k<=n;k++)
                {
                    if(map[j][k]=='1')
                        indegree[k]--;
                }
                break;
            }
        }
        if(!flag) break;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int g=1;g<=t;g++)
    {
        scanf("%d",&n);
        memset(indegree,0,sizeof(indegree));
        for(int i=1;i<=n;i++)
        {
            scanf("%s",map[i]+1);
            for(int j=1;j<=n;j++)
                if(map[i][j]=='1')
                    indegree[j]++;
        }
        cnt=0;
        toposort();
        cout<<"Case #"<<g<<": ";
        if(cnt==n) cout<<"No"<<endl;
        else cout<<"Yes"<<endl;
    }


    return 0;
}