同学可以开到 5e3 + 5 AC , 我开到 101 MLE ; 求助

源代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <queue>

//#define int long long
#define INF 0x3f3f3f
#define inf -0x3f3f3f
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

using namespace std ;

int n , m ;

bool f[101][101] ;

int ans ;

inline void read(int &x) {
	bool ok ; char ch ;
	for(ok = false , ch = getchar() ; !isdigit(ch) ; ch = getchar()) {
		if(ch == '-') ok = true ;
	}
	for(x = 0 ; isdigit(ch) ; x = x * 10 + ch - '0' , ch = getchar()) {
		if(ok) x = -x ; 
	}
}

inline void write(int x) {
	if(n < 0) {
		putchar('-') ;
		x *= -x ;
	}
	if(n > 9) write(x / 10) ;
	putchar(x % 10 + '0') ;
}

signed main ( ) {
	
	read(n) , read(m) ;
	for(int i = 1 ; i <= m ; i ++) {
		int x , y ;
		read(x) , read(y) ;
		f[x][y] = true ;
	}
	
	for(int k = 1 ; k <= n ; k ++) {
		for(int i = 1 ; i <= n ; i ++) {
			for(int j = 1 ; j <= n ; j ++) {
				f[i][j] = f[i][j] or (f[i][k] and f[k][j]) ;
			}
		}
	}
	
	for(int i = 1 ; i <= n ; i ++) {
		bool sum = 1 ;
		for(int j = 1 ; j <= n ; j ++) {
			if(i == j) continue ;
			else sum = sum and (f[i][j] or f[j][i]) ;
		}
		ans += sum ;
	}
	
	write(ans) ;
	putchar('\n') ;
	
	return 0 ;
}