首页 > 代码库 > HDU 4324:Triangle LOVE( 拓扑排序 )

HDU 4324:Triangle LOVE( 拓扑排序 )

2024-07-12 16:58:00   220人阅读

Triangle LOVE



Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2271    Accepted Submission(s): 946


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). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,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
2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110
 

Sample Output
Case #1: Yes
Case #2: No
 
题意:给你一个特殊的有向图,该有向图的任意两个节点u与v之间有且仅有一条单向边,现在问你该有向图是否存在由3个节点构成的环.

该图本质是拓扑排序题.如果该图可以拓扑排序,那么不存在3节点的环,否则存在3节点的环.



#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>

using namespace std;

const int M = 2000 + 5;
int n;
int in[M];
char str[M];
int t;
vector<int> map[M];

bool toposort()
{
    int sum = 0;
    queue<int>Q;
    for(int i=0; i<n; i++)
        if( !in[i] )
        Q.push( i );
    while( !Q.empty() )
    {
        int u = Q.front();
        Q.pop();
        sum++;
        for(int i=0; i<map[u].size(); i++)
        {
            int m = map[u][i];
            if( --in[m] == 0 )
                Q.push( m );
        }
    }
    if( sum==n )
        return true;
    else
        return false;
}

int main()
{
    scanf( "%d", &t );
    int cas;
    for( cas=1; cas<=t; cas++ )
    {
        scanf( "%d", &n );
        memset( in, 0, sizeof( in ) );
        for( int i=0; i<n; i++ )
        {
            map[i].clear();
            scanf( "%s", str );
            for( int j=0; j<n; j++ )
            //for(int j=0; j<strlen(str); j++)  
            //这么写会超时，复杂度会增加
            {
                if( str[j]=='1' )
                   {
                       map[i].push_back( j );
                       in[ j ]++;
                   }
            }
        }
         printf("Case #%d: %s\n", cas, toposort()?"No":"Yes");
    }

    return 0;
}








联系
我们