洛谷P4281 [AHOI2008]紧急集合 / 聚会

(Large extbf{Description: } large{一棵n个节点的树，m次操作，每次给出三个节点x,y,z，求三个节点到树上某个节点距离之和的\最小值，并输出此节点。(1 leq x,y,z leq n leq 5 imes 10^{5}, 1 leq m leq 5 imes 10^{5})}\)

(Large extbf{Solution: } large{我们很容易的想到 ext{LCA}，显然三个节点两两之间的 ext{LCA}一定有一个是同一个节点，那么我们\要求的节点就是此点，因为可证无论是这个节点祖先，儿子还是兄弟都不可能更优。\对于距离，我们可以在 ext{dfs}的时候顺便处理，然后用前缀和的思想O(1)查询就好了。})

(Large extbf{Code: })

#include <cstdio>
#include <algorithm>
#define LL long long
#define gc() getchar()
#define rep(i, a, b) for(int i = (a); i <= (b); ++i)
#define _rep(i, a, b) for(int i = (a); i >= (b); --i)
using namespace std;
const int N = 5e5 + 5;
int n, m, cnt, head[N], size[N], son[N], top[N], fa[N], dep[N];

struct Edge {
	int to, next;
}e[N << 1];

inline int read() {
	char ch = gc();
	int ans = 0, flag = 1;
	while (ch > '9' || ch < '0') ch = gc();
	while (ch >= '0' && ch <= '9') ans = (ans << 1) + (ans << 3) + ch - '0', ch = gc();
	return ans * flag;
}

inline void add(int x, int y) {
	e[++cnt].to = y;
	e[cnt].next = head[x];
	head[x] = cnt;	
}

inline void dfs1(int x, int y) {
	fa[x] = y;
	dep[x] = dep[y] + 1;
	size[x] = 1;
	int Max = 0;
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == y) continue;
		dfs1(u, x);
		size[x] += size[u];
		if (size[u] > Max) son[x] = u, Max = size[u];
	}
}

inline void dfs2(int x, int tp) {
	top[x] = tp;
	if (!son[x]) return;
	dfs2(son[x], tp);
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == son[x] || u == fa[x]) continue;
		dfs2(u, u);
	}	
}

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

int main() {
	n = read(), m = read();
	int x, y, z;
	rep(i, 2, n) x = read(), y = read(), add(x, y), add(y, x);
	dfs1(1, 0);
	dfs2(1, 1);
	while (m--) {
		x = read(), y = read(), z = read();
		int f = lca(x, y), ff = lca(x, z), fff = lca(y, z);
		if (dep[f] == dep[ff]) printf("%d ", fff);
		else if (dep[ff] == dep[fff]) printf("%d ", f);
		else printf("%d ", ff);
		printf("%d
", dep[x] + dep[y] + dep[z] - min(dep[f], min(dep[ff], dep[fff])) * 2 - max(dep[f], max(dep[ff], dep[fff])));
	} 
	return 0;
}