线段树合并RE on #1求调
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线段树合并RE on #1求调
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TankYu楼主2024/9/5 13:56

RT

#include <iostream>
#include <vector>
#define LL long long
using namespace std;
int n, q;

vector<int> e[300010];
int dep[300010];
struct node
{
    LL sum;
    int ls;
    int rs;
} t[300010 << 5];
LL siz[300010];
void pushup(int rt)
{
    t[rt].sum = t[t[rt].ls].sum + t[t[rt].rs].sum;
    return;
}
int TOT;
void update(int &rt, int l, int r, int x, int k)
{
    if (!rt)
    {
        rt = ++TOT;
    }
    t[rt].sum += k;
    if (l == r)
    {
        return;
    }
    int mid = (l + r) / 2;
    if (x <= mid)
    {
        update(t[rt].ls, l, mid, x, k);
    }
    else
    {
        update(t[rt].rs, mid + 1, r, x, k);
    }
    return;
}

int merge(int x, int y)
{
    if (!x || !y)
    {
        return x + y;
    }
    int rt = ++TOT;
    t[rt].sum = t[x].sum + t[y].sum;
    t[rt].ls = merge(t[x].ls, t[y].ls);
    t[rt].rs = merge(t[x].rs, t[y].rs);
    return rt;
}

LL query(int rt, int l, int r, int ql, int qr)
{
    if (rt == 0)
    {
        return 0;
    }
    if (ql <= l && r <= qr)
    {
        return t[rt].sum;
    }
    int mid = (l + r) / 2;
    LL ans = 0;
    if (ql <= mid)
    {
        ans += query(t[rt].ls, l, mid, ql, qr);
    }
    if (qr > mid)
    {
        ans += query(t[rt].rs, mid + 1, r, ql, qr);
    }
    return ans;
}
int rt[300010];
void dfs(int x, int fa)
{
    siz[x] = 1;
    dep[x] = dep[fa] + 1;
    for (auto i : e[x])
    {
        if (i == fa)
        {
            continue;
        }
        dfs(i, x);
        siz[x] += siz[i];
    }
    // update(1, siz, );
    update(rt[x], 1, n, dep[x], siz[x] - 1);
    if (fa != 0)
    {
        rt[fa] = merge(rt[fa], rt[x]);
    }
    return;
}

int main()
{
    cin >> n >> q;
    for (int i = 1; i < n; i++)
    {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs(1, 0);
    while (q--)
    {
        int a, k;
        cin >> a >> k;
        cout << min(dep[a] - 1, k) * (siz[a] - 1) + query(rt[a], 1, n, dep[a] + 1, min(n, dep[a] + k)) << '\n';
    }
    return 0;
}
2024/9/5 13:56
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