【文件夹合并——树链剖分，树状数组】

题目

复杂度

代码

#pragma GCC optimize(3)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const int N = 5e5 + 10;
const int M = 1e6 + 10;

int n, m;
vector<int> sub[N];                                           // 子文件夹
ll d[N];                                                      // 数据量
int h[N], e[M], ne[M], idx;                                   // 链式前向星
int dep[N], sz[N], son[N], dfn[N], fa[N], top[N], id[N], cnt; // 树链剖分
int tr[N];                                                    // 树状数组
void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void dfs1(int u)
{
    sz[u] = 1;
    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (j == fa[u])
            continue; // fa[root] = 0;
        dep[j] = dep[u] + 1;
        fa[j] = u;
        dfs1(j);
        sz[u] += sz[j];
        if (sz[j] > sz[son[u]])
            son[u] = j;
    }
}

void dfs2(int u, int t)
{
    dfn[u] = ++cnt;
    id[cnt] = u;
    top[u] = t;

    if (!son[u])
        return;
    else
        dfs2(son[u], t);

    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (j == fa[u] || j == son[u])
            continue;
        dfs2(j, j);
    }
}

ll query(int x)
{
    ll retv = 0;

    for (; x; x -= x & -x)
        retv += tr[x];

    return retv;
}

ll query(int l, int r)
{
    return query(r) - query(l - 1);
}
void modify(int x, int v)
{
    for (; x <= n; x += x & -x)
        tr[x] += v;
}

ll nquery(int a, int b)
{
    ll retv = 0;

    while (top[a] != top[b])
    {
        if (dep[top[a]] < dep[top[b]])
            swap(a, b);
        retv += query(dfn[top[a]], dfn[a]);
        a = fa[top[a]];
    }

    if (dep[a] > dep[b])
        swap(a, b);
    retv += query(dfn[a], dfn[b]);
    
    return retv;
}

void nmodify(int x, int v)
{
    modify(dfn[x], v);
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    memset(h, -1, sizeof h);
    
    cin >> n >> m;
    for (int i = 2; i <= n; i++)
    {
        int f;
        cin >> f;
        sub[f].push_back(i);
        add(f, i);
        add(i, f);
    }
    for (int i = 1; i <= n; i++)
        cin >> d[i];

    dfs1(1);
    dfs2(1, 1);
    
    for(int i = 1; i <= n; i++)
        nmodify(i, 1);
        
    while (m--)
    {
        int op, x;
        cin >> op >> x;

        if (op == 1)
        {
            vector<int> t;
            for (int i : sub[x])
            {
                d[x] += d[i];
                nmodify(i, -1);
                for (int j : sub[i])
                    t.push_back(j);
            }

            sub[x] = t;
            cout << sub[x].size() << ' ' << d[x] << '\n';
        }
        else if (op == 2)
            cout << nquery(1, x) << '\n';
    }
}