树句节狗提 [树状数组 / (dep, dfn)二维数点]

$树句节狗提$

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$color{red}{正解部分}$

先题意转换:

求 子树内 所有与 $x$ 距离 至少为 $K$ 的点的权值和

$downarrow$

子树内 的和 $-$ 子树内 与 $x$ 距离小于等于$K-1$ 的点的权值和.

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$color{red}{实现部分}$

树状数组 中下标为 深度, 按 $dfs$序 顺序将点权加入 树状数组, 保证 深度不超过 $k-1$ 这个条件,

把询问 离线, 转化为差分操作, 按$dfs$序遍历到子树的根时减, 离开子树时再加, 可以得到 子树内深度不超过 $K-1$ 的值 .

• 离线可以使用 $vector$ 或者 链表 .
• 在差分右端点时注意边界 !!!

#include<bits/stdc++.h>
#define fi first
#define se second
#define reg register
#define pb push_back
typedef long long ll;
typedef std::pair<int, int> pr;

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 2525015;

int N;
int Q_;
int dft;
int num0;
int A[maxn];
int Mp[maxn];
int dep[maxn];
int dfn[maxn];
int head[maxn];
int sum_n[maxn];

ll Ans[maxn];

std::vector <pr> que[maxn];

struct Edge{ int nxt, to; } edge[maxn << 1];

void Add(int from, int to){
        edge[++ num0] = (Edge){ head[from], to };
        head[from] = num0;
}

struct Bit_Tree{
        ll v[maxn];
        void Add(int k, int p){ while(k<=N)v[k]+=p,k+=k&-k; }
        ll Query(int k){ ll s = 0; while(k)s+=v[k],k-=k&-k; return s; }
} bit_1;

void DFS(int k, int fa){
        dfn[k] = ++ dft, sum_n[k] = 1;
        dep[k] = dep[fa] + 1, Mp[dft] = k;
        for(reg int i = head[k]; i; i = edge[i].nxt){
                int to = edge[i].to;
                if(to == fa) continue ;
                DFS(to, k), sum_n[k] += sum_n[to];
        }
}

void Calc_ans(){
        ll Tmp_1 = 0;
        for(reg int i = 1; i <= N; i ++){
                int x = Mp[i];
                bit_1.Add(dep[x], A[x]);
                Tmp_1 += A[x];
                int size = que[i].size();
                for(reg int j = 0; j < size; j ++){
                        int q_id = que[i][j].fi, dep_k = que[i][j].se;
                        if(q_id >= 0) Ans[q_id] += Tmp_1 - bit_1.Query(dep_k);
                        else Ans[-q_id] += bit_1.Query(dep_k) - Tmp_1;
                }
        }
}

void print(int q, long long* ans, int lim);

int main(){
        N = read();
        for(reg int i = 1; i <= N; i ++) A[i] = read();
        for(reg int i = 2; i <= N; i ++){
                int x = read();
                Add(x, i), Add(i, x);
        }
        DFS(1, 0); Q_ = read();
        for(reg int i = 1; i <= Q_; i ++){
                int x = read(), k = read();
                k = std::min(N, dep[x]+k-1);
                que[dfn[x]-1].pb(pr(-i, k));
                que[dfn[x]+sum_n[x]-1].pb(pr(i, k));
        }
        Calc_ans();
        print(Q_, Ans, read());
        return 0;
}

void print(int q, long long* ans, int lim) {
        for(int i = 1; i <= q; ) {
                long long res = 0;
                for(int j = i; j <= std::min(q, i + lim - 1); j ++) res ^= ans[j];
                i += lim;
                printf("%lld
", res);
        }
        return ;
}