[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=2836
[算法]
树链剖分
时间复杂度 : O(NlogN ^ 2)
[代码]
#include<bits/stdc++.h> using namespace std; #define MAXN 100010 typedef long long LL; struct edge { int to , nxt; } e[MAXN << 1]; int n , tot , timer; int head[MAXN] , size[MAXN] , son[MAXN] , fa[MAXN] , dfn[MAXN] , depth[MAXN] , top[MAXN]; template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); } template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0'; x *= f; } struct Segment_Tree { struct Node { int l , r; LL sum , tag; } Tree[MAXN << 2]; inline void build(int index , int l , int r) { Tree[index].l = l; Tree[index].r = r; Tree[index].tag = Tree[index].sum = 0; if (l == r) return; int mid = (l + r) >> 1; build(index << 1 , l , mid); build(index << 1 | 1 , mid + 1 , r); } inline void pushdown(int index) { int l = Tree[index].l , r = Tree[index].r; int mid = (l + r) >> 1; Tree[index << 1].sum += (mid - l + 1) * Tree[index].tag; Tree[index << 1 | 1].sum += (r - mid) * Tree[index].tag; Tree[index << 1].tag += Tree[index].tag; Tree[index << 1 | 1].tag += Tree[index].tag; Tree[index].tag = 0; } inline void update(int index) { Tree[index].sum = Tree[index << 1].sum + Tree[index << 1 | 1].sum; } inline void modify(int index , int l , int r , LL value) { if (Tree[index].l == l && Tree[index].r == r) { Tree[index].sum += value * (r - l + 1); Tree[index].tag += value; return; } pushdown(index); int mid = (Tree[index].l + Tree[index].r) >> 1; if (mid >= r) modify(index << 1 , l , r , value); else if (mid + 1 <= l) modify(index << 1 | 1 , l , r , value); else { modify(index << 1 , l , mid , value); modify(index << 1 | 1 , mid + 1 , r , value); } update(index); } inline LL query(int index , int l , int r) { if (Tree[index].l == l && Tree[index].r == r) return Tree[index].sum; pushdown(index); int mid = (Tree[index].l + Tree[index].r) >> 1; if (mid >= r) return query(index << 1 , l , r); else if (mid + 1 <= l) return query(index << 1 | 1 , l , r); else return query(index << 1 , l , mid) + query(index << 1 | 1 , mid + 1 , r); } } SGT; inline void addedge(int u , int v) { ++tot; e[tot] = (edge){v , head[u]}; head[u] = tot; } inline void dfs1(int u) { son[u] = -1; size[u] = 1; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v == fa[u]) continue; fa[v] = u; depth[v] = depth[u] + 1; dfs1(v); size[u] += size[v]; if (son[u] == -1 || size[v] > size[son[u]]) son[u] = v; } } inline void dfs2(int u , int tp) { dfn[u] = ++timer; top[u] = tp; if (son[u] != -1) dfs2(son[u] , tp); for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v == fa[u] || v == son[u]) continue; dfs2(v , v); } } inline void modify(int u , int v , LL d) { int tu = top[u] , tv = top[v]; while (tu != tv) { if (depth[tu] > depth[tv]) { swap(u , v); swap(tu , tv); } SGT.modify(1 , dfn[tv] , dfn[v] , d); v = fa[tv]; tv = top[v]; } if (depth[u] > depth[v]) swap(u , v); SGT.modify(1, dfn[u] , dfn[v] , d); } int main() { scanf("%d" , &n); for (int i = 1; i < n; i++) { int u , v; scanf("%d%d" , &u , &v); addedge(u , v); fa[v] = u; } dfs1(0); dfs2(0 , 0); SGT.build(1 , 1 , n); int q; scanf("%d" , &q); while (q--) { char op[5]; scanf("%s" , &op); if (op[0] == 'A') { int u , v; LL d; scanf("%d%d%lld" , &u , &v , &d); modify(u , v , d); } else { int u; scanf("%d" , &u); printf("%lld\n" , SGT.query(1 , dfn[u] , dfn[u] + size[u] - 1)); } } return 0; }