P2853. Cow Picnic S

考点

  • DFS

题解

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#include <bits/stdc++.h>
using namespace std;
const int LEN = 1e3 + 50;
int n, m, k, tot, ans, head[LEN], vis[LEN], cow[LEN], cnt[LEN];
struct edge {
int to_, nxt_;
} e[(int)1e5 + 50];

void add(int u, int v) {
e[tot].nxt_ = head[u], e[tot].to_ = v;
head[u] = tot++;
}

void dfs(int x) {
if (vis[x]) return;
vis[x] = 1;
++cnt[x];
for (int i = head[x]; ~i; i = e[i].nxt_) {
dfs(e[i].to_);
}
}

int main() {
int u, v;
cin >> k >> n >> m;
memset(head, -1, sizeof(head));
for (int i = 0; i < k; ++i) cin >> cow[i];
for (int i = 0; i < m; ++i) {
cin >> u >> v;
add(u, v);
}
for (int i = 0; i < k; ++i) {
memset(vis, 0, sizeof(vis));
dfs(cow[i]);
}
for (int i = 1; i <= n; ++i) {
if (cnt[i] == k) ++ans;
}
cout << ans;
return 0;
}

思路

从奶牛所在牧场开始dfs,所经路径中的牧场的被访问次数加一

dfs结束后,遍历所有牧场的被访问次数,等于k就是我们需要的