题面
解法
动态点分治模板题
什么是动态点分治呢？？？
静态的点分治就是不断地找到当前树的重心，然后分成若干个子树继续递归下去
但是如果有修改似乎静态的就不好处理了
我们现在引入一个叫点分树的东西
说白了，就是每一次静态点分治的时候这一次的重心和上一次的连边，这样形成了一棵树
再说清楚一点，就是如果这棵树上有一条边\((x,y)\)，\(x\)是\(y\)的父亲
那么说明\(x\)是某一棵子树的重心，\(y\)是\(x\)子树的重心
根据重心的性质，这一棵树的深度不超过\(O(log\ n)\)
对于这道题，修改时就是在点分树上找到这个点，然后不断向上，用线段树维护区间加法
查询的时候不断向上爬，累加答案
注意处理点分树上父亲对儿子答案产生的影响
时间复杂度：\(O(n\ log^2\ n)\)
代码
#include#define inf INT_MAX#define N 100010using namespace std;template void chkmax(node &x, node y) {x = max(x, y);}template void chkmin(node &x, node y) {x = min(x, y);}template void read(node &x) { x = 0; int f = 1; char c = getchar(); while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();} while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;}int n, q, rt, now, cnt, d[N], p[N], ff[N], vis[N], siz[N], f[N][18];struct Edge { int next, num;} e[N * 3];struct SegmentTree { struct Node { int lc, rc, del; } t[N * 100]; int tot, rt[N]; void New(int &k) {if (!k) k = ++tot;} void modify(int &k, int l, int r, int L, int R, int v) { if (L > R) return; New(k); if (L <= l && r <= R) { t[k].del += v; return; } int mid = (l + r) >> 1; if (L <= mid && mid < R) modify(t[k].lc, l, mid, L, mid, v), modify(t[k].rc, mid + 1, r, mid + 1, R, v); if (R <= mid) modify(t[k].lc, l, mid, L, R, v); if (L > mid) modify(t[k].rc, mid + 1, r, L, R, v); } void Add(int x, int l, int r, int v) {modify(rt[x], 0, n, l, r, v);} int query(int k, int l, int r, int x) { if (!k) return 0; if (l == r) return t[k].del; int ans = t[k].del; int mid = (l + r) >> 1; if (x <= mid) return ans + query(t[k].lc, l, mid, x); return ans + query(t[k].rc, mid + 1, r, x); } int ask(int x, int d) {return query(rt[x], 0, n, d);}} T1, T2;void add(int x, int y) { e[++cnt] = (Edge) {e[x].next, y}; e[x].next = cnt;}void getr(int x, int fa) { siz[x] = 1, ff[x] = 0; for (int p = e[x].next; p; p = e[p].next) { int k = e[p].num; if (k == fa || vis[k]) continue; getr(k, x); siz[x] += siz[k]; ff[x] = max(ff[x], siz[k]); } ff[x] = max(ff[x], now - siz[x]); if (ff[x] < ff[rt]) rt = x;}void work(int x, int fa) { vis[x] = 1, p[x] = fa; for (int p = e[x].next; p; p = e[p].next) { int k = e[p].num; if (k == fa || vis[k]) continue; ff[rt = 0] = inf, now = siz[k]; getr(k, x); work(rt, x); }}void dfs(int x, int fa) { d[x] = d[fa] + 1; for (int i = 1; i <= 17; i++) f[x][i] = f[f[x][i - 1]][i - 1]; for (int p = e[x].next; p; p = e[p].next) { int k = e[p].num; if (k == fa) continue; f[k][0] = x; dfs(k, x); }}int lca(int x, int y) { if (d[x] < d[y]) swap(x, y); for (int i = 17; i >= 0; i--) if (d[f[x][i]] >= d[y]) x = f[x][i]; if (x == y) return x; for (int i = 17; i >= 0; i--) if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i]; return f[x][0];}int dis(int x, int y) { int z = lca(x, y); return d[x] + d[y] - 2 * d[z];}void Modify(int x, int d, int v) { T1.Add(x, 0, d, v); for (int y = x; p[y]; y = p[y]) { int t = dis(x, p[y]); T1.Add(p[y], 0, d - t, v); T2.Add(y, 0, d - t, v); }}int Query(int x) { int ret = T1.ask(x, 0); for (int y = x; p[y]; y = p[y]) { int t = dis(p[y], x); ret += T1.ask(p[y], t) - T2.ask(y, t); } return ret;}int main() { read(n), read(q); cnt = n; for (int i = 1; i < n; i++) { int x, y; read(x), read(y); add(x, y), add(y, x); } ff[rt = 0] = inf, now = n; getr(1, 0), work(rt, 0), dfs(1, 0); while (q--) { char c = getchar(); while (!isalpha(c)) c = getchar(); if (c == 'M') { int x, d, v; read(x), read(d), read(v); Modify(x, d, v); } else { int x; read(x); cout << Query(x) << "\n"; } } return 0;}