P4281 [ahoi2008] emergency assembly / gathering (LCA)

[AHOI2008] Emergency assembly / party - Luogu https://www.luogu.com.cn/problem/P4281 This question obviously needs to be used LCA Write , seek 3 One point, two liang LCA And then determine , If this 3 individual LCA Equal, then they meet at that point , If one is different , At that different meeting , It is impossible to have three different situations .

Correctness can be judged by drawing a picture .

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cstring>
#include <set>
#include <cmath>
#include <map>
#include <bitset>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int MN = 65005;
const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
#define IOS ios::sync_with_stdio(false)
#define lowbit(x) ((x)&(-x))
using P = pair<int, int>;



int n, m;
int head[MAXN];
int nxt[MAXN];
int ver[MAXN];
int cnt;
int dep[MAXN];
int LOG2[MAXN];
int fafa[MAXN][20];
void add(int x, int y) {
	ver[++cnt] = y;
	nxt[cnt] = head[x];
	head[x] = cnt;
}

void dfs(int p, int fa) {
	fafa[p][0] = fa;
	for (int i = 1; i <= LOG2[dep[p]]; i++) {
		fafa[p][i] = fafa[fafa[p][i - 1]][i - 1];
	}
	for (int i = head[p]; i; i = nxt[i]) {
		int v = ver[i];
		if (v == fa)
			continue;
		dep[v] = dep[p] + 1;
		dfs(v, p);
	}
}

int lca(int x, int y) {
	if (dep[x] < dep[y])
		swap(x, y);
	while (dep[x] != dep[y]) {
		x = fafa[x][LOG2[dep[x] - dep[y]]];
	}
	if (x == y)
		return x;
	for (int i = LOG2[dep[x]]; i >= 0; i--) {
		if (fafa[x][i] != fafa[y][i])
			x = fafa[x][i], y = fafa[y][i];
	}
	return fafa[x][0];
}

void solve() {
	int x, y, z;
	scanf("%d %d %d", &x, &y, &z);
	int tmp1 = lca(x, y);
	int tmp2 = lca(y, z);
	int tmp3 = lca(x, z);
//	printf("====%d %d %d\n", tmp1, tmp2, tmp3);
	if (tmp1 == tmp2 && tmp1 == tmp3 && tmp2 == tmp3) {
		printf("%d %d\n", tmp1, dep[x] + dep[y] + dep[z] - 3 * dep[tmp1]);
	} else if (tmp1 == tmp2 && tmp1 != tmp3) {
		printf("%d %d\n", tmp3, dep[x] + dep[z] - 2 * dep[tmp3] + dep[tmp3] + dep[y] - 2 * dep[tmp1]);
	} else if (tmp1 == tmp3 && tmp1 != tmp2) {
		printf("%d %d\n", tmp2, dep[y] + dep[z] - 2 * dep[tmp2] + dep[tmp2] + dep[x] - 2 * dep[tmp1]);
	} else if (tmp2 == tmp3 && tmp1 != tmp2) {
		printf("%d %d\n", tmp1, dep[x] + dep[y] - 2 * dep[tmp1] + dep[tmp1] + dep[z] - 2 * dep[tmp2]);
	}
}


int main() {
	scanf("%d %d", &n, &m);
	for (int i = 2; i <= n; i++) {
		LOG2[i] = LOG2[i / 2] + 1;
	}
	dep[1] = 0;
	int x, y;
	for (int i = 1; i <= n - 1; i++) {
		scanf("%d %d", &x, &y);
		add(x, y);
		add(y, x);
	}
	dfs(1, 0);
	while (m--)
		solve();
	return 0;
}