洛谷 P3833 [AHOI2012] 树上的操作

Fri, 14 Aug 2026 23:24:19 +0800

给群里红石镐的题解备份，免得翻记录翻两百条。

题意

一棵树，支持两种操作：

路径加：把 $u$ 到 $v$ 路径上每个点权值加 $w$
区间和：求 $u$ 到 $v$ 路径上的点权和

思路

板子题，树链剖分 + 线段树。两遍 DFS 剖出重链，把树上问题变成区间问题，再用线段树维护 dfn 序。

代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 100005;

int n, m;
int head[N], to[N << 1], nxt[N << 1], etot;
int fa[N], dep[N], siz[N], son[N], top[N], dfn[N], tot;
ll sum[N << 2], lazy[N << 2];

void add(int u, int v) {
 to[++etot] = v;
 nxt[etot] = head[u];
 head[u] = etot;
}

// 第一遍 DFS：求 fa, dep, siz, son
void dfs1(int u, int f) {
 fa[u] = f;
 dep[u] = dep[f] + 1;
 siz[u] = 1;
 for (int e = head[u]; e; e = nxt[e]) {
 int v = to[e];
 if (v == f) continue;
 dfs1(v, u);
 siz[u] += siz[v];
 if (siz[v] > siz[son[u]]) son[u] = v;
 }
}

// 第二遍 DFS：剖重链，求 top, dfn
void dfs2(int u, int t) {
 top[u] = t;
 dfn[u] = ++tot;
 if (!son[u]) return;
 dfs2(son[u], t);
 for (int e = head[u]; e; e = nxt[e]) {
 int v = to[e];
 if (v == fa[u] || v == son[u]) continue;
 dfs2(v, v);
 }
}

void pushup(int p) {
 sum[p] = sum[p << 1] + sum[p << 1 | 1];
}

void pushdown(int p, int l, int r) {
 if (lazy[p]) {
 int mid = (l + r) >> 1;
 lazy[p << 1] += lazy[p];
 lazy[p << 1 | 1] += lazy[p];
 sum[p << 1] += lazy[p] * (mid - l + 1);
 sum[p << 1 | 1] += lazy[p] * (r - mid);
 lazy[p] = 0;
 }
}

void build(int p, int l, int r) {
 if (l == r) {
 sum[p] = 0;
 return;
 }
 int mid = (l + r) >> 1;
 build(p << 1, l, mid);
 build(p << 1 | 1, mid + 1, r);
 pushup(p);
}

void update(int p, int l, int r, int ql, int qr, ll w) {
 if (ql <= l && r <= qr) {
 sum[p] += w * (r - l + 1);
 lazy[p] += w;
 return;
 }
 pushdown(p, l, r);
 int mid = (l + r) >> 1;
 if (ql <= mid) update(p << 1, l, mid, ql, qr, w);
 if (qr > mid) update(p << 1 | 1, mid + 1, r, ql, qr, w);
 pushup(p);
}

ll query(int p, int l, int r, int ql, int qr) {
 if (ql <= l && r <= qr) return sum[p];
 pushdown(p, l, r);
 int mid = (l + r) >> 1;
 ll res = 0;
 if (ql <= mid) res += query(p << 1, l, mid, ql, qr);
 if (qr > mid) res += query(p << 1 | 1, mid + 1, r, ql, qr);
 return res;
}

void upd_path(int u, int v, ll w) {
 while (top[u] != top[v]) {
 if (dep[top[u]] < dep[top[v]]) swap(u, v);
 update(1, 1, n, dfn[top[u]], dfn[u], w);
 u = fa[top[u]];
 }
 if (dep[u] > dep[v]) swap(u, v);
 update(1, 1, n, dfn[u], dfn[v], w);
}

ll qry_path(int u, int v) {
 ll res = 0;
 while (top[u] != top[v]) {
 if (dep[top[u]] < dep[top[v]]) swap(u, v);
 res += query(1, 1, n, dfn[top[u]], dfn[u]);
 u = fa[top[u]];
 }
 if (dep[u] > dep[v]) swap(u, v);
 res += query(1, 1, n, dfn[u], dfn[v]);
 return res;
}

int main() {
 ios::sync_with_stdio(false);
 cin.tie(nullptr);
 cin >> n >> m;
 for (int i = 1; i < n; i++) {
 int x, y;
 cin >> x >> y;
 add(x, y);
 add(y, x);
 }
 dfs1(1, 0);
 dfs2(1, 1);
 build(1, 1, n);
 while (m--) {
 string op;
 cin >> op;
 if (op == "Add") {
 int u, v;
 ll w;
 cin >> u >> v >> w;
 upd_path(u, v, w);
 } else if (op == "Query") {
 int u, v;
 cin >> u >> v;
 cout << qry_path(u, v) << '\n';
 }
 }
 return 0;
}

算法 on 羽衣甘蓝的博客

洛谷 P3833 [AHOI2012] 树上的操作

题意

思路

代码