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.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;
}
HDU 4324:Triangle LOVE( 拓扑排序 ),布布扣,bubuko.com
HDU 4324:Triangle LOVE( 拓扑排序 )
原文：http://blog.csdn.net/u013487051/article/details/37914423