CodeForces 1827E Bus Routes

洛谷传送门

CF 传送门

比较神奇的题。

定一个非叶子 \(r\) 为根。

显然只用判断两个叶子是否可达。

求出每个叶子向上能一步跳到的深度最浅的点 \(p_i\)，那么如果 \(p_i\) 不在一条链上就无解，因为两条路径没有交点。

然后只用判断 \(p_i\) 最深的叶子的 \(p_i\) 能不能一步到达其他叶子即可。

code

// Problem: E. Bus Routes
// Contest: Codeforces - Codeforces Round 873 (Div. 1)
// URL: https://codeforces.com/problemset/problem/1827/E
// Memory Limit: 1024 MB
// Time Limit: 2500 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;

const int maxn = 500100;

int n, m, head[maxn], len, deg[maxn], b[maxn];
int fa[maxn], dep[maxn], son[maxn], sz[maxn], top[maxn];
bool vis[maxn];
pii a[maxn];
vector<int> vc[maxn];

struct edge {
	int to, next;
} edges[maxn << 1];

inline void add_edge(int u, int v) {
	edges[++len].to = v;
	edges[len].next = head[u];
	head[u] = len;
}

int dfs(int u, int f, int d) {
	fa[u] = f;
	sz[u] = 1;
	dep[u] = d;
	int maxson = -1;
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (v == f) {
			continue;
		}
		sz[u] += dfs(v, u, d + 1);
		if (sz[v] > maxson) {
			son[u] = v;
			maxson = sz[v];
		}
	}
	return sz[u];
}

void dfs2(int u, int tp) {
	top[u] = tp;
	vis[u] = 1;
	if (!son[u]) {
		return;
	}
	dfs2(son[u], tp);
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (!vis[v]) {
			dfs2(v, v);
		}
	}
}

inline int qlca(int x, int y) {
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) {
			swap(x, y);
		}
		x = fa[top[x]];
	}
	if (dep[x] > dep[y]) {
		swap(x, y);
	}
	return x;
}

void solve() {
	len = 0;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		head[i] = deg[i] = son[i] = 0;
		vector<int>().swap(vc[i]);
		vis[i] = 0;
	}
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v);
		add_edge(u, v);
		add_edge(v, u);
		++deg[u];
		++deg[v];
	}
	for (int i = 1; i <= m; ++i) {
		scanf("%d%d", &a[i].fst, &a[i].scd);
		vc[a[i].fst].pb(a[i].scd);
		vc[a[i].scd].pb(a[i].fst);
	}
	if (n == 2) {
		for (int i = 1; i <= m; ++i) {
			if (a[i].fst != a[i].scd) {
				puts("YES");
				return;
			}
		}
		puts("NO\n1 2");
		return;
	}
	int r = -1;
	for (int i = 1; i <= n; ++i) {
		if (deg[i] > 1) {
			r = i;
			break;
		}
	}
	dfs(r, -1, 1);
	dfs2(r, r);
	int p = -1, q = -1, mx = -1e9;
	vector<pii> ps;
	for (int u = 1; u <= n; ++u) {
		if (deg[u] != 1) {
			continue;
		}
		int k = -1, mn = 1e9;
		if (vc[u].empty()) {
			printf("NO\n%d %d\n", u, r);
			return;
		}
		for (int v : vc[u]) {
			int w = qlca(u, v);
			if (dep[w] < mn) {
				mn = dep[w];
				k = w;
			}
		}
		b[u] = k;
		ps.pb(k, u);
		if (dep[k] > mx) {
			mx = dep[k];
			p = k;
			q = u;
		}
	}
	sort(ps.begin(), ps.end(), [&](pii x, pii y) {
		return dep[x.fst] < dep[y.fst];
	});
	for (int i = 0; i + 1 < (int)ps.size(); ++i) {
		if (qlca(ps[i].fst, ps[i + 1].fst) != ps[i].fst) {
			printf("NO\n%d %d\n", ps[i].scd, ps[i + 1].scd);
			return;
		}
	}
	for (int u = 1; u <= n; ++u) {
		if (deg[u] != 1 || u == q) {
			continue;
		}
		bool flag = 1;
		for (int v : vc[u]) {
			if (qlca(v, p) == p || qlca(u, p) == p) {
				flag = 0;
				break;
			}
		}
		if (flag) {
			printf("NO\n%d %d\n", u, q);
			return;
		}
	}
	puts("YES");
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}