Zijian-lv #3 树句节狗提

如你所见，这是一道狗题

一棵树，多次询问与一个点距离至少为 $k$ 的点的权值和

$n,q leq 2525010$

sol：

长链剖分

需要注意的是这道题卡空间

我把我所有的 vector 换成链表才过了

#include <bits/stdc++.h>
#define LL long long
#define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i)
#define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i)
using namespace std;
inline int read() {
    int x = 0, f = 1; char ch;
    for (ch = getchar(); !isdigit(ch); ch = getchar()) if (ch == '-') f = -f;
    for (; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0';
    return x * f;
}
const int maxn = 2525015;
int n, q, lim;/*
vector<pair<int, int> > qs[maxn];
vector<int> G[maxn];*/
int qcd[maxn];
struct Chain {
    int head[maxn], nx[maxn], id[maxn], sz;
    Chain() {memset(head, 0, sizeof(head)); sz = 0;}
    inline void AddItem(int pos, int Item, int other = -1) {
        id[++sz] = Item;
        nx[sz] = head[pos];
        head[pos] = sz;
        if(~other) qcd[sz] = other;
    }
}qs, G;
LL ans[maxn], pool[maxn], *f[maxn], *now=pool+1;
int a[maxn];
void print(int q, LL* ans, int lim) {
    LL res; 
    for(int i = 1; i <= q; ) {
        res = 0;
        for(int j = i; j <= min(q, i + lim - 1); j++) res ^= ans[j];
        i += lim;
        printf("%lld
", res);
    }
    return ;
}
int mxs[maxn], mxd[maxn];
void dfs(int x) {
    mxd[x] = 0;
    for(int i=G.head[x];i;i=G.nx[i]) {
        dfs(G.id[i]);
        mxd[x] = max(mxd[x], mxd[G.id[i]] + 1);
        if((mxd[G.id[i]] > mxd[mxs[x]])) mxs[x] = G.id[i];
    }
}
void solve(int x) {
    if(mxs[x]) {
        f[mxs[x]] = f[x] + 1;
        solve(mxs[x]);
    }
    f[x][0] = a[x];
    for(int i=G.head[x];i;i=G.nx[i]) {
        if(G.id[i] == mxs[x]) continue;
        f[G.id[i]] = now; now += mxd[G.id[i]] + 1; solve(G.id[i]);
        rep(j, 0, mxd[G.id[i]]) f[x][j+1] += f[G.id[i]][j];
        //cout << "to: " << to << " " << f[x][1] << endl; 
    }
    //cout << x << " " << f[x][1] << endl;
    /*dwn(i, mxd[x]-1, 0) {
        cout << f[x][i] << " jiale " << f[x][i + 1] << endl;
        f[x][i] += f[x][i + 1];
    }*/
    f[x][0] += f[x][1];
    for(int i=qs.head[x];i;i=qs.nx[i]) {
        if(qs.id[i] > mxd[x]) ans[qcd[i]] = 0;
        else ans[qcd[i]] = f[x][qs.id[i]];
    }/*
    if(qs[x].size()) {
    //    cout << "DEBUG: " << qs[x].first << " " << qs[x].second << " " << f[x][qs[x].first] << endl;
    //    if(qs[x].first > mxd[x]) ans[qs[x].second] = 0;
    //    else
        /*cout << "x: " << x << endl;
        rep(i, 0, mxd[x])
            cout << f[x][i] << " ";
        cout << endl;*/
    /*    
        for(auto ii : qs[x]) {
            //cout << "DEBUG: " << ii.first << " " << mxd[x] << endl;
            if(ii.first > mxd[x]) ans[ii.second] = 0;
            else ans[ii.second] = f[x][ii.first];
        }
    }*/
}
int main() {
    n = read(); int x, y;
    rep(i, 1, n) a[i] = read();
    rep(i, 2, n) {
        //x = read(), G[x].push_back(i);
        x = read();
        G.AddItem(x, i);
    }
    q = read();
    rep(i, 1, q) {
        x = read(); y = read();
        //qs[x].push_back(make_pair(y, i));
        qs.AddItem(x, y, i);
    } mxd[0] = -1;
    dfs(1);
    f[1] = now; now += mxd[1] + 1; solve(1);
    lim = read();
    print(q, ans, lim);
    //rep(i, 1, q) cout << ans[i] << endl;
}

 论我的没 log 跑得没别人带 log 快