【POJ】3660 Cow Contest floyd（可以拓扑排序？）

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
USACO 2008 January Silver

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 ;
}