3730: 震波

3730: 震波

链接

分析：

　　动态点分治。

　　求距离小于等于k的点权和。

　　建出点分树，然后对于每个分治中心，维护连通块到这个点的所有距离，因为要容斥掉多计算的，所以在维护这个点到这个分治中心在点分树的父节点的距离。

　　动态开点线段树，下标为距离，记录权值和。

　　空间复杂福：$nlog^2n$，时间复杂度$nlog^2n$。加上大常数后，跑的有点慢，加上fread后，14904ms卡过。

代码：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<bitset>
#define fore(i, u, v) for (int i = head[u], v = e[i].to; i; i = e[i].nxt, v = e[i].to) 
using namespace std;
typedef long long LL;

char buf[100000], *p1 = buf, *p2 = buf;
#define nc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
inline int read() {
    int x=0,f=1;char ch=nc();for(;!isdigit(ch);ch=nc())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=nc())x=x*10+ch-'0';return x*f;
}

const int N = 100005;
struct Edge { int to, nxt; } e[N << 1];
int head[N], dep[N], f[N << 1][20], Log[N << 1], pos[N], siz[N], val[N], fa[N];
int En, Index, TreeIndex, Mn, Root, n;
bool vis[N];
int sum[N * 200], ls[N * 200], rs[N * 200], Ra[N], Rb[N];

inline void add_edge(int u,int v) {
    ++En; e[En].to = v, e[En].nxt = head[u]; head[u] = En;
    ++En; e[En].to = u, e[En].nxt = head[v]; head[v] = En;
}
void update(int l,int r,int &now,int p,int v) {
    if (!now) now = ++TreeIndex;
    sum[now] += v;
    if (l == r) return ;
    int mid = (l + r) >> 1;
    if (p <= mid) update(l, mid, ls[now], p, v);
    else update(mid + 1, r, rs[now], p, v);
}
int query(int l,int r,int now,int p) {
    if (!now) return 0;
    if (l == r) return sum[now];
    int mid = (l + r) >> 1;
    if (p <= mid) return query(l, mid, ls[now], p);
    else return query(mid + 1, r, rs[now], p) + sum[ls[now]];
}
void predfs(int u,int fa) {
    pos[u] = ++Index; f[Index][0] = dep[u];
    fore (i, u, v) if (v != fa) dep[v] = dep[u] + 1, predfs(v, u), f[++Index][0] = dep[u];
}
void prermq() {
    for (int i = 2; i <= Index; ++i) Log[i] = Log[i >> 1] + 1;
    for (int j = 1; j <= Log[Index]; ++j)
        for (int i = 1; i + (1 << j) - 1 <= Index; ++i)
            f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
}
int LCA(int x,int y) {
    x = pos[x], y = pos[y];
    if (x > y) swap(x, y);
    int k = Log[y - x + 1];
    return min(f[x][k], f[y - (1 << k) + 1][k]);
}
int getdis(int x,int y) { return dep[x] + dep[y] - 2 * LCA(x, y); }

void getroot(int u,int fa,int Size) {
    int mx = 0; siz[u] = 1;
    fore (i, u, v)
        if (!vis[v] && v != fa) 
            getroot(v, u, Size), siz[u] += siz[v], mx = max(mx, siz[v]);
    mx = max(mx, Size - siz[u]);
    if (mx < Mn) Mn = mx, Root = u;
}
void work(int u,int pre,int r) {
    update(0, n - 1, Ra[r], getdis(u, r), val[u]); 
    if (fa[r]) update(0, n - 1, Rb[r], getdis(u, fa[r]), val[u]);
    siz[u] = 1;
    fore (i, u, v) if (!vis[v] && v != pre) work(v, u, r), siz[u] += siz[v];
}
void solve(int u) {
    work(u, 0, u); vis[u] = 1; 
    fore (i, u, v) if (!vis[v]) Mn = 1e9, getroot(v, u, siz[v]), fa[Root] = u, solve(Root);    
}
void Change(int x,int y) {
    for (int i = x; i; i = fa[i]) {
        update(0, n - 1, Ra[i], getdis(x, i), y - val[x]);
        if (fa[i]) update(0, n - 1, Rb[i], getdis(x, fa[i]), y - val[x]);
    }
    val[x] = y;
}
int Ask(int x,int k) {
    LL ans = 0;
    for (int i = x; i; i = fa[i]) {
        if (getdis(i, x) <= k) ans += query(0, n - 1, Ra[i], k - getdis(x, i));
        if (fa[i] && getdis(x, fa[i]) <= k) ans -= query(0, n - 1, Rb[i], k - getdis(x, fa[i]));
    }
    return ans;
}
int main() {
    n = read();int  m = read();
    for (int i = 1; i <= n; ++i) val[i] = read();
    for (int i = 1; i < n; ++i) add_edge(read(), read());
    predfs(1, 0); prermq();
    Mn = 1e9, getroot(1, 0, n);
    solve(Root);     
    int opt, x, y, lastans = 0;
    while (m --) {
        opt = read(), x = read() ^ lastans, y = read() ^ lastans;
        if (opt) Change(x, y);
        else printf("%d\n", lastans = Ask(x, y));
    }
    return 0;
}