BZOJ 4448。

寫起來很愉悅的題。

按照時間可持久化線段樹，修改就在原來的地方加$1$即可，查詢就直接詢問$root_1 - root_{now - c - 1}$中相應的個數。

主席樹維護樹鏈剖分即可。

時間複雜度$O(nlog^2n)$。

Code：

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 2e5 + 5;

int n, rt, qn, tot = 0, head[N];
int dfsc = 0, id[N], siz[N], fa[N], top[N], son[N], dep[N];

 
struct Edge {
    int to, nxt;
} e[N << 1];

inline void add(int from, int to) {
    e[++tot].to = to;
    e[tot].nxt = head[from];
    head[from] = tot;
}

inline void read(int &X) {
    X = 0; char ch = 0; int op = 1;
    for(; ch > '9' || ch < '0'; ch = getchar())
        
 
if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

inline void swap(int &x, int &y) {
    int t = x; x = y; y = t;
}
 
void dfs1(int x, int fat, int depth) {
    fa[x] = fat, dep[x] = depth, siz[x] = 1
 
;
    int maxson = -1;
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fat) continue;
        dfs1(y, x, depth + 1);

        siz[x] += siz[y];
        if(siz[y] > maxson) {
            son[x] = y;
            maxson = siz[y];
        }
    }
}

void dfs2(int x, int topf) {
    top[x] = topf, id[x] = ++dfsc;
    if(!son[x]) return;
    dfs2(son[x], topf);
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fa[x] || y == son[x]) continue;
        dfs2(y, y);
    }
}

namespace PSegT {
    struct Node {
        int lc, rc, sum;
    } s[N * 40];

    int root[N], nodeCnt = 0;

    #define lc(p) s[p].lc
    #define rc(p) s[p].rc
    #define mid ((l + r) >> 1)
    #define sum(p) s[p].sum

    void ins(int &p, int l, int r, int x, int pre) {
        s[p = ++nodeCnt] = s[pre];
        ++sum(p);
        if(l == r) return;

        if(x <= mid) ins(lc(p), l, mid, x, lc(pre));
        else ins(rc(p), mid + 1, r, x, rc(pre));
    }

    int query(int r1, int r2, int l, int r, int x, int y) {
        if(x <= l && y >= r) return sum(r2) - sum(r1);

        int res = 0;
        if(x <= mid) res += query(lc(r1), lc(r2), l, mid, x, y);
        if(y > mid) res += query(rc(r1), rc(r2), mid + 1, r, x, y);

        return res;
    }

} using namespace PSegT;

inline void solve(int r) {
    int x, y, c, res = 0;
    read(x), read(y), read(c);
    r -= c + 1;
    int px = x, py = y;

    for(; top[x] != top[y]; ) {
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        if(r > 0) res += query(root[0], root[r], 1, n, id[top[x]], id[x]);
        x = fa[top[x]];
    }

    if(dep[x] > dep[y]) swap(x, y);
    if(r > 0) res += query(root[0], root[r], 1, n, id[x], id[y]);

    printf("%d %d\n", dep[px] + dep[py] - 2 * dep[x] + 1, res);
}

int main() {
    read(n);
    for(int fat, i = 1; i <= n; i++) {
        read(fat);
        if(!fat) rt = i;
        else add(i, fat), add(fat, i);
    }
    dfs1(rt, 0, 1), dfs2(rt, rt);

    read(qn);
    for(int op, i = 1; i <= qn; i++) {
        root[i] = root[i - 1];
        read(op);
        if(op == 2) {
            int x; read(x);
            ins(root[i], 1, n, id[x], root[i]);
        } else solve(i);
    }
    
    return 0;
}