HDU 4324 拓扑排序-优快云博客

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1387 Accepted Submission(s): 584

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 
  

    针对一个 
   顶点活动网(Activity On Vertex network)，简称 
   AOV网，从中去除入度为0的顶点，同时更新从改点出发引起的入度，让这些点的入度减1，直到最后如果AOV网为空时，说明那么去除的这些点就组成了一个拓扑排序，如果AOV网不为空，这种情况若在程序中出现，则称为死锁或死循环，是应该必须避免的，说明这些活动是永远执行不到的。（活动的前驱又是在活动之后执行） 
  

    AC代码: 
  

 
     #include <stdio.h>
#include <string.h>
int t,n;
//存储的是节点的入度
int in_degree[2013];
//存储的是i,j两个结点的关系，1:i love j,0:j love i
char adj_mat[2013][2013];

int main()
{
    bool flag;//true表示为有三角恋，false表示为没有三角恋
    scanf("%d",&t);
    for(int i = 1; i <= t;i++)
    {

        scanf("%d",&n);
        flag = false;
        //将所有的结点入度初始化为0
        memset(in_degree,0,sizeof(in_degree));
        for(int j = 0; j < n; j++)
        {
            scanf("%s",adj_mat[j]);
            for(int k=0;k<n;k++)
            if(adj_mat[j][k]=='1')//如果j喜欢k,则把k的入度加1
            in_degree[k]++;
        }

        for(j=0;j<n;j++)
        {
            int k;
            for(k=0;k<n;k++)
            if(in_degree[k]==0)break;//找出入度为0的节点结点
            if(k==n)//任何一个结点的入度都不为0，说明存在环了，则必有三角恋
            {
                flag = true;
                break;
            }else{
                //将这个点的入度设为-1，避免再次循环时又查到了这个结点，
                //此时说明这个点已经从集合中除掉了
                in_degree[k]--;
                for(int p=0;p<n;p++)
                {
                    //把从这个节点出发的引起的结点的入度都减去1
                    if(adj_mat[k][p]=='1'&&in_degree[p]!=0)
                    in_degree[p]--;
                }
            }
        }
        if(flag)
        printf("Case #%d: Yes\n",i);
        else printf("Case #%d: No\n",i);
    }
    return 0;
} 
    
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