064:Cow Contest

064:Cow Contest（弗洛伊德模板題）

思路：

以一個點爲跳板，比較他已明確的兩個點之間的關係，一共存在兩種關係：
1：i>k&&k>j then i>j
2：i<k&&k<j then i<j
暴力跑就完事了

代碼：

#include<iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <functional>
#include <ctime>
#include <iomanip>
#include <sstream>
#include <algorithm>
#define ll long long
#define mes(x,y) memset(x,y,sizeof(x))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
const ll INF = 1e9 + 10;
ll dp[500][500];
ll n, m, k, i, j, t, x, y;
ll Floyd() {
	for (k = 1; k <= n; k++) {
		for (i = 1; i <= n; i++) {
			if (dp[i][k] == 0)continue;
			for (j = 1; j <= n; j++) {
				if (dp[i][k] == dp[k][j]) {
					dp[i][j] = dp[i][k];
				}
			}
		}
	}
	ll sum = 0;
	for (i = 1; i <= n; i++) {
		ll ans = 0;
		for (j = 1; j <= n; j++) {
			if (dp[i][j])ans++;
		}
		if (ans == n - 1)sum++;
	}
	return sum;
}
int main() {
	FAST_IO;
	while (cin >> n >> m) {
		mes(dp, 0);
		for (i = 0; i < m; i++) {
			cin >> x >> y;
			dp[x][y] = 1;
			dp[y][x] = -1;
		}
		cout << Floyd() << endl;
	}
}