P2853 [USACO06DEC] Cow Picnic S

简单的一道深搜：
思路：让每个有奶牛的农场沿着道路跑下去，跑到的农场加上root地方的奶牛数量，最后统计能有k头奶牛的农场数量就是答案
代码：

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
typedef long long ll;
using namespace std;
const int N = 1e3 + 10;
ll hascow[N];
ll numofcow[N];
bool vis[N];
ll ans = 0;
vector<int>G[N];
void dfs(ll u,ll root)
{
	vis[u] = 1;
	for (int i = 0; i < G[u].size(); i++)
	{
		if(!vis[G[u][i]])
		{
			numofcow[G[u][i]] += hascow[root];
			dfs(G[u][i], root);
		}
	}
	
}
int main()
{
	ios::sync_with_stdio(false); cin.tie(0), cout.tie(0);
	ll k, n, m; cin >> k >> n >> m;
	queue<ll>qid;
	for (int i = 0; i < k; i++)
	{
		ll x; cin >> x;
		if(!hascow[x])qid.push(x);
		hascow[x] += 1;
		numofcow[x] = hascow[x];
	}
	for (int i = 0; i < m; i++)
	{
		ll u, v; cin >> u >> v;
		G[u].push_back(v);
	}
	//思路：从有牛的节点找起，然后给所有这个牛能到的节点加1，统计满的节点数量
	while (!qid.empty())
	{
		dfs(qid.front(),qid.front());
		qid.pop();
		memset(vis, 0, sizeof(vis));
	}
	for (int i = 1; i <= n; i++)
		if (numofcow[i] == k)
			ans++;
	cout << ans;
	return 0;
}

很神奇：那个一段如果改为以下：

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
typedef long long ll;
using namespace std;
const int N = 1e3 + 10;
ll hascow[N];
ll numofcow[N];
bool vis[N];
ll ans = 0;
vector<int>G[N];
void dfs(ll u,ll root)
{
	vis[u] = 1;
	for (int i = 0; i < G[u].size(); i++)
	{
		if(!vis[G[u][i]])
		{
			numofcow[G[u][i]] += hascow[root];
			dfs(G[u][i], root);
		}
	}
	vis[u] = 0;
}
int main()
{
	ios::sync_with_stdio(false); cin.tie(0), cout.tie(0);
	ll k, n, m; cin >> k >> n >> m;
	queue<ll>qid;
	for (int i = 0; i < k; i++)
	{
		ll x; cin >> x;
		if(!hascow[x])qid.push(x);
		hascow[x] += 1;
		numofcow[x] = hascow[x];
	}
	for (int i = 0; i < m; i++)
	{
		ll u, v; cin >> u >> v;
		G[u].push_back(v);
	}
	//思路：从有牛的节点找起，然后给所有这个牛能到的节点加1，统计满的节点数量
	while (!qid.empty())
	{
		dfs(qid.front(),qid.front());
		qid.pop();
	}
	for (int i = 1; i <= n; i++)
		if (numofcow[i] == k)
			ans++;
	cout << ans;
	return 0;
}

就过不了
待查为什么...