Triangle LOVE

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

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

Case #1: Yes
Case #2: No

该图本质是拓扑排序题.假设该图能够拓扑排序,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，复杂度会添加
            {
                if( str[j]=='1' )
                   {
                       map[i].push_back( j );
                       in[ j ]++;
                   }
            }
        }
         printf("Case #%d: %s\n", cas, toposort()?"No":"Yes");
    }

    return 0;
}