题目链接:http://poj.org/problem?id=3660

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 ≤ AN; 1 ≤ BN; AB), 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

题目大意:有N头牛,评以N个等级,各不相同,先给出部分牛的等级的高低关系,问最多能确定多少头牛的等级
解题思路:一头牛的等级,当且仅当它与其它N-1头牛的关系确定时确定,于是我们可以将牛的等级关系看做一张图,然后进行适当的松弛操作,得到任意两点的关系,再对没一头牛进行检查即可
 #include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm> using namespace std; int map[][], INF = 0x3f3f3f3f; int main(){
ios::sync_with_stdio( false ); int n, m;
cin >> n >> m;
memset( map, INF, sizeof( map ) ); int x, y;
for( int i = ; i < m; i++ ){
cin >> x >> y;
map[x][y] = ; //x战胜y
map[y][x] = -; //y败于x
} for( int j = ; j <= n; j++ )
for( int i = ; i <= n; i++ )
for( int k = ; k <= n; k++ ){
if( map[i][j] == map[j][k] && ( map[i][j] == || map[i][j] == - ) ) //进行松弛
map[i][k] = map[i][j];
} int ans = ;
for( int i = ; i <= n; i++ ){
int sum = ;
for( int j = ; j <= n; j++ ){
if( map[i][j] != INF )
sum++;
}
if( sum == n - )
ans++;
} cout << ans << endl; return ;
}
												

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