洛谷P2590 [ZJOI2008]树的统计 题解 树链剖分+线段树

题目链接：https://www.luogu.org/problem/P2590
树链剖分模板题。
剖分过程要用到如下7个值：

• fa[u]：u的父节点编号；
• dep[u]：u的深度；
• size[u]：u为根的子树中节点总数；
• son[u]：u的重儿子；
• top[u]：u所在的重链的顶部节点；
• seg[u]：u在线段树中的位置；
• rev[u]：seg的倒置，即rev[seg[u]] == u。

然后套线段树模板实现区间最值、区间和，及单点更新操作。

实现代码如下：

#include <bits/stdc++.h>
using namespace std;
#define INF (1<<29)
const int maxn = 30030;
int fa[maxn],
	dep[maxn],
	size[maxn],
	son[maxn],
	top[maxn],
	seg[maxn], seg_cnt,
	rev[maxn],
	n, w[maxn], maxv[maxn<<2], sumv[maxn<<2];
vector<int> g[maxn];
void dfs1(int u, int p) {
	size[u] = 1;
	for (vector<int>::iterator it = g[u].begin(); it != g[u].end(); it ++) {
		int v = (*it);
		if (v == p) continue;
		fa[v] = u;
		dep[v] = dep[u] + 1;
		dfs1(v, u);
		size[u] += size[v];
		if (size[v] >size[son[u]]) son[u] = v;
	}
}
void dfs2(int u, int tp) {
	seg[u] = ++seg_cnt;
	rev[seg_cnt] = u;
	top[u] = tp;
	if (son[u]) dfs2(son[u], tp);
	for (vector<int>::iterator it = g[u].begin(); it != g[u].end(); it ++) {
		int v = (*it);
		if (v == fa[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}
#define lson l, mid, rt<<1
#define rson mid+1, r, rt<<1|1
void push_up(int rt) {
	sumv[rt] = sumv[rt<<1] +sumv[rt<<1|1];
	maxv[rt] = max(maxv[rt<<1], maxv[rt<<1|1]);
}
void build(int l, int r, int rt) {
	int mid = (l + r) / 2;
	if (l == r) {
		sumv[rt] = maxv[rt] = w[rev[l]];
		return;
	}
	build(lson); build(rson);
	push_up(rt);
}
void update(int p, int v, int l, int r, int rt) {
	if (l == r) {
		sumv[rt] = maxv[rt] = v;
		return;
	}
	int mid = (l + r) / 2;
	if (p <= mid) update(p, v, lson);
	else update(p, v, rson);
	push_up(rt);
}
int query_max(int L, int R, int l, int r, int rt) {
	if (L <= l && r <= R) return maxv[rt];
	int mid = (l + r) / 2, tmp = -INF;
	if (L <= mid) tmp = max(tmp, query_max(L, R, lson));
	if (R > mid) tmp = max(tmp, query_max(L, R, rson));
	return tmp;
}
int query_sum(int L, int R, int l, int r, int rt) {
	if (L <= l && r <= R) return sumv[rt];
	int mid = (l + r) / 2, tmp = 0;
	if (L <= mid) tmp += query_sum(L, R, lson);
	if (R > mid) tmp += query_sum(L, R, rson);
	return tmp;
}
int ask_max(int u, int v) {
	int res = -INF;
	while (top[u] != top[v]) {
		if (dep[top[u]] < dep[top[v]]) swap(u, v);
		res = max(res, query_max(seg[top[u]], seg[u], 1, n, 1));
		u = fa[top[u]];
	}
	if (dep[u] < dep[v]) swap(u, v);
	res = max(res, query_max(seg[v], seg[u], 1, n, 1));
	return res;
}
int ask_sum(int u, int v) {
	int res = 0;
	while (top[u] != top[v]) {
		if (dep[top[u]] < dep[top[v]]) swap(u, v);
		res += query_sum(seg[top[u]], seg[u], 1, n, 1);
		u = fa[top[u]];
	}
	if (dep[u] < dep[v]) swap(u, v);
	res += query_sum(seg[v], seg[u], 1, n, 1);
	return res;
}
int m;
string s;
int main() {
	cin >> n;
	for (int i = 1; i < n; i ++) {
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	for (int i = 1; i <= n; i ++) cin >> w[i];
	dep[1] = fa[1] = 1;
	dfs1(1, -1);
	dfs2(1, 1);
	build(1, n, 1);
	cin >> m;
	while (m --) {
		int u, v;
		cin >> s >> u >> v;
		if (s == "CHANGE") update(seg[u], v, 1, n, 1);
		else if (s == "QMAX") cout << ask_max(u, v) << endl;
		else cout << ask_sum(u, v) << endl;
	}
	return 0;
}