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



[align=left]Problem Description[/align]
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”.

[align=left]Input[/align]
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).

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Case #1: Yes
Case #2: No

题意:给你一个特殊的有向图,该有向图的随意两个节点u与v之间有且仅有一条单向边,如今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.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;
}