聚会

题目描述(https://loj.ac/problem/10136)

思路

3个节点a，b，c选择（a,b)(a,c)(b,c）公共祖先深度最大的节点，3个节点的最大距离为每个节点到公共祖先的距离和 / 2；

代码

#include <cstdio>
#include <algorithm>

const int MAX = 5e5 + 5;
int n, m;
int head[MAX], ver[MAX << 1], nt[MAX << 1], ht;
int f[MAX][21], dep[MAX];
int ans, res, dps;

inline int read() {
	int s = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') ch = getchar();
	while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s;
}

void add(int x, int y) {
	nt[++ht] = head[x], head[x] = ht, ver[ht] = y;
}
void dfs_lca(int x, int y) {
	dep[x] = dep[y] + 1;
	f[x][0] = y;
	for (int i = 1; i <= 19; ++i) f[x][i] = f[f[x][i - 1]][i - 1];
	for (int i = head[x], j; i; i = nt[i]) {
		j = ver[i];
		if (j == y) continue;
		dfs_lca(j, x);
	}
}
int lca(int x, int y) {
	if (dep[x] < dep[y]) std::swap(x, y);
	for (int i = 19; i >= 0; --i) {
		if (dep[f[x][i]] >= dep[y]) {
			x = f[x][i];
			ans += (1 << i);
		}
	}
	if (x == y) {
		if (dps < dep[x]) res = x, dps = dep[x];
		return x;
	}
	for (int i = 19; i >= 0; --i) {
		if (f[x][i] != f[y][i]) {
			x = f[x][i], y = f[y][i];
			ans += 1 << (i + 1);
		}
	}
	ans += 2;
	if (dps < dep[f[x][0]]) res = f[x][0], dps = dep[f[x][0]];
	return f[x][0];
}
int main() {
	n = read(), m = read();
	for (int i = 1, a, b; i < n; ++i) {
		a = read(), b = read();
		add(a, b), add(b, a);
	}
	dfs_lca(1, 0);
	for (int i = 1, a, b, c; i <= m; ++i) {
		a = read(), b = read(), c = read(), ans = 0, dps = 0;
		int ab = lca(a, b), ac = lca(a, c), bc = lca(b, c);
		// printf("lca:%d %d %d
", ab, ac, bc);
		printf("%d %d
", res, ans >> 1);
	}
	return 0;
}