【POJ】3660 Cow Contest floyd（可以拓扑排序？）

Cow Contest

Time Limit: 1000MS
Memory Limit: 65536KTotal 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 cowA has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠ B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (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 andM
* 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 54 34 23 21 22 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 ;}