hdoj 4324 Triangle LOVE (拓扑排序)

最新推荐文章于 2020-05-06 14:04:37 发布

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本文介绍了一个有趣的问题——如何通过编程检测一组人中是否存在三角恋关系。利用拓扑排序的方法，可以有效地判断给定的关系矩阵中是否存在这样的三角关系。

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Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4662 Accepted Submission(s): 1833

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

判断是否有向图成环，用拓扑排序

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int map[2010][2010];
int in[2010];
int n,k;
void topo()
{
    int ans=0;
    queue<int> q;
    for(int i=0;i<n;i++)
    {
        if(in[i]==0)
            q.push(i);
    }
    while(!q.empty())
    {
        int a=q.front();
        in[a]--;
        ans++;
        q.pop();
        for(int j=0;j<n;j++)
        {
            if(map[a][j])
            {
                in[j]--;
                if(in[j]==0)
                    q.push(j);
            }
        }
    }
    if(ans!=n)
        printf("Case #%d: Yes\n",k++);
    else
        printf("Case #%d: No\n",k++);
}
int main()
{
    int t,x;
    char str[2010];
    k=1;
    scanf("%d",&t);
    while(t--)
    {
        memset(map,0,sizeof(map));
        memset(in,0,sizeof(in));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%s",str);
            for(int j=0;j<n;j++)
            {            
                if(str[j]=='1')
                {
                    map[i][j]=1;
                    in[j]++;
                }
            }
        }
        topo();
    }
    return 0;
}