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Cow Contest
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6925 | Accepted: 3792 |
Description
N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.
The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.
Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B
Output
* Line 1: A single integer representing the number of cows whose ranks can be determined
Sample Input
5 5
4 3
4 2
3 2
1 2
2 5
Sample Output
2
Source
传送门：【POJ】3660 Cow Contest floyd
题目分析：floyd求传递闭包，然后对每个点数比他大的个数和比他小的个数，如果和等于n-1，说明该点的名次可以确定。
Orz风神，我做的是floyd。。。可是这题在拓扑排序题集里面。。。不愧是刷光图论的神。。根本想不到拓扑的做法。T U T
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REPV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define clear( a , x ) memset ( a , x , sizeof a )typedef long long Int ; const int MAXN = 105 ;
const int INF = 0x3f3f3f3f ;int G[MAXN][MAXN] ;
int win[MAXN] , lose[MAXN] ;
int n , m ;void work () {int u , v ;while ( ~scanf ( "%d%d" , &n , &m ) ) {clear ( G , 0 ) ;clear ( win , 0 ) ;clear ( lose , 0 ) ;REP ( i , m ) {scanf ( "%d%d" , &u , &v ) ;G[u][v] = 1 ;}REPF ( k , 1 , n )REPF ( i , 1 , n )REPF ( j , 1 , n )G[i][j] |= G[i][k] & G[k][j] ;REPF ( i , 1 , n )REPF ( j , 1 , n )if ( G[i][j] )++ win[i] , ++ lose[j] ;int cnt = 0 ;REPF ( i , 1 , n )if ( win[i] + lose[i] + 1 == n )++ cnt ;printf ( "%d\n" , cnt ) ;}
}int main () {work () ;return 0 ;
}
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