「CF1037D」Valid BFS?

传送门
Luogu

解题思路

考虑直接模拟 ( ext{BFS}) 的过程。
对于每一个节点的儿子，先遍历在输入序列中靠前的，判断 ( ext{BFS}) 是否匹配即可。

细节注意事项

• 注意一下输出格式

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cmath>
#include <ctime>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while (!isdigit(c)) f |= (c == '-'), c = getchar();
	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
	s = f ? -s : s;
}

const int _ = 200010;

int n, in[_], p[_];
int hd, tl, q[_], vis[_];
vector < int > G[_];

inline bool cmp(const int& x, const int& y) { return p[x] < p[y]; }

inline void bfs() {
	hd = tl = 0, q[++tl] = 1;
	while (hd < tl) {
		int u = q[++hd], X = G[u].size();
		vis[u] = 1;
		for (rg int i = 0; i < X; ++i) {
			int v = G[u][i];
			if (!vis[v]) q[++tl] = v;
		}
	}
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
#endif
	read(n);
	for (rg int u, v, i = 1; i < n; ++i) read(u), read(v), G[u].push_back(v), G[v].push_back(u);
	for (rg int i = 1; i <= n; ++i) read(in[i]), p[in[i]] = i;
	for (rg int i = 1; i <= n; ++i) sort(G[i].begin(), G[i].end(), cmp);
	bfs();
	for (rg int i = 1; i <= n; ++i) if (in[i] != q[i]) return puts("No"), 0;
	puts("Yes");
	return 0;
}

完结撒花 (qwq)