POJ 3660 Cow Contest floyd dfs

題意：有n個奶牛，給出m個奶牛的關係，即某個奶牛的編程能力比另一個要強，求出能夠確定排名的奶牛的最大數量
思路：建立n個點m條邊的有向圖，某個點能夠確定排名，則其他所有點不是可以到達該點，就是可以由該點到達。可以用floyd來判斷，給每條邊權值賦1，如果松弛後某點與其他所有點的邊權值不爲無窮大，則該點就能確定排名

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<stack> 
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e2+10;
const int inf = 0x3f3f3f3f;
int n, m, g[maxn][maxn];
int main()
{
	while (cin >> n >> m) {
		memset(g, inf, sizeof(g));
		 for (int i = 0; i < m; i++) {
		 	int x, y;
		 	cin >> x >> y;
		 	g[x][y] = 1;
		 }
		 for (int k = 1; k <= n; k++)
		 	for (int i = 1; i <= n; i++)
		 		for (int j = 1; j <= n; j++)
		 			if (g[i][j] > g[i][k] + g[k][j])
		 				g[i][j] = g[i][k] + g[k][j];
		int ans = 0;
		for (int i = 1; i <= n; i++) {
			int cnt = 0;
			for (int j = 1; j <= n; j++) {
				if (i != j && (g[i][j] != inf || g[j][i] != inf))
					cnt++;
			}
			if (cnt == n-1)
				ans++;
		}
		cout << ans << "\n";
	} 
	return 0;
}

思路2：用兩遍dfs，第二遍將邊反向，統計一個點能夠被到達的次數，若該點在兩次dfs後能被其他所有點到達，則該點能確定排名

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<stack> 
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e2+10;
const int inf = 0x3f3f3f3f;
int n, m, g[maxn][maxn], d[maxn], vis[maxn];
void dfs(int s)
{
	vis[s] = 1;
	for (int j = 1; j <= n; j++) {
		if (j != s && g[s][j] && !vis[j]) {
			dfs(j);
			d[j]++;
		}
	}
}
int main()
{
	while (cin >> n >> m) {
		 memset(g, 0, sizeof(g));
	 	 memset(d, 0, sizeof(d));
		 for (int i = 0; i < m; i++) {
		 	int x, y;
		 	cin >> x >> y;
		 	g[x][y] = 1;
		 }
		 for (int i = 1; i <= n; i++) {
		 	memset(vis, 0, sizeof(vis));
		 	dfs(i);
		 }
		 for (int i = 1; i <= n; i++)
		 	for (int j = 1; j < i; j++) {
		 		swap(g[i][j], g[j][i]);
		 	}
		 for (int i = 1; i <= n; i++) {
		 	memset(vis, 0, sizeof(vis));
		 	dfs(i);
		 }
		 int ans = 0;
		 for (int i = 1; i <= n; i++) {
		 	if (d[i] == n-1)
		 		ans++;
		 }
		 cout << ans << "\n";
	} 
	return 0;
}