HDU 4324:Triangle LOVE( 拓扑排序 )

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). 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

 

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;}