BZOJ3730 震波 | 动态点分治

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <vector>
#define space putchar(' ')
#define enter putchar('\n')
using namespace std;
typedef long long ll;
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
    if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
    x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}

const int N = 100005, INF = 0x3f3f3f3f;
int n, m, ans;
int ecnt, adj[N], nxt[2*N], go[2*N];
int val[N], fa[N][20], dis[N][20], dep[N], sze[N], son[N], rt, sz;
bool vis[N];
vector <int> bit[N], fbit[N];

/* 变量解释:
fa[i][j]  : 第j次分治中, i所属的连通块的重心
dis[i][j] : i到fa[i][j]的距离
dep[i]    : 点分树上i的深度
sze, son, rt, sz 用于求重心
vis 用于点分治标注
bit[i] : 当i为重心时, 能求"离i距离为j的点的权值和是多少"的树状数组
fbit[i]: 当fa[i][dep[i] - 1]为重心时, 能求"含i连通块离fa[i][dep[i] - 1]距离为j的点的权值和是多少"的树状数组
*/

void add(int u, int v){
    go[++ecnt] = v;
    nxt[ecnt] = adj[u];
    adj[u] = ecnt;
}
void getG(int u, int pre){
    sze[u] = 1, son[u] = 0;
    for(int e = adj[u], v; e; e = nxt[e])
	if(!vis[v = go[e]] && v != pre){
	    getG(v, u);
	    sze[u] += sze[v];
	    son[u] = max(son[u], sze[v]);
	}
    son[u] = max(son[u], sz - sze[u]);
    if(son[u] < son[rt]) rt = u;
}
void dfs(int u, int pre, int top, int d){
    for(int e = adj[u], v; e; e = nxt[e])
	if(!vis[v = go[e]] && v != pre){
	    fa[v][++dep[v]] = top;
	    dis[v][dep[v]] = d;
	    dfs(v, u, top, d + 1);
	}
}
void build(int u){
    vis[u] = 1;
    dfs(u, 0, u, 1);
    int all = sz;
    bit[u].resize(all + 1);
    fbit[u].resize(all + 1);
    for(int e = adj[u], v; e; e = nxt[e])
	if(!vis[v = go[e]]){
	    sz = sze[v] > sze[u] ? all - sze[u] : sze[v];
	    rt = 0;
	    getG(v, u);
	    build(rt);
	}
}
int ask(int u, int k){
    int ret = val[u], lim = bit[u].size() - 1;
    for(k = min(k, lim); k; k -= k & -k) ret += bit[u][k];
    return ret;
}
int fask(int u, int k){
    int ret = 0, lim = fbit[u].size() - 1;
    for(k = min(k, lim); k; k -= k & -k) ret += fbit[u][k];
    return ret;
}
void change(int u, int x){
    int lim = bit[u].size() - 1;
    for(int j = dis[u][dep[u]]; j <= lim && j; j += j & -j) fbit[u][j] += x;
    for(int i = dep[u]; i; i--){
	lim = bit[fa[u][i]].size() - 1;
	for(int j = dis[u][i]; j <= lim; j += j & -j) bit[fa[u][i]][j] += x;
	for(int j = dis[u][i - 1]; j <= lim && j; j += j & -j) fbit[fa[u][i]][j] += x;
    }
}
int query(int u, int k){
    int ret = ask(u, k);
    for(int i = dep[u]; i; i--)
	if(dis[u][i] <= k)
	    ret += ask(fa[u][i], k - dis[u][i]) - fask(fa[u][i + 1], k - dis[u][i]);
    return ret;
}
int main(){

    read(n), read(m);
    for(int i = 1; i <= n; i++) read(val[i]);
    for(int i = 1, u, v; i < n; i++)
	read(u), read(v), add(u, v), add(v, u);
    son[0] = INF, sz = n;
    getG(1, 0);
    build(rt);
    for(int i = 1; i <= n; i++) fa[i][dep[i] + 1] = i;
    for(int i = 1; i <= n; i++) change(i, val[i]);
    int op, x, y;
    while(m--){
	read(op), read(x), read(y);
	x ^= ans, y ^= ans;
	//printf("op = %d, x = %d, y = %d\n", op, x, y);
	if(op == 0) write(ans = query(x, y)), enter;
	else change(x, y - val[x]), val[x] = y;
    }

    return 0;
}