2018 Multi-University Training Contest 7

LCT + 树上倍增

和bzoj的弹飞绵羊差不多，多了一个倍增记录父亲节点。

这里有个小技巧，把飞出去的点当成第1个比较好操作。。不像弹飞绵羊是第n+1个，因为这里是往上跳，所以出去了的话我们的父亲节点就是0，直接把节点+1当初LCT中的点就好了

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 100005;
int _, n, m, t, tot, ch[N][2], size[N], rev[N], fa[N], p[N][21], st[N], to[N];

void build(){
    ++ tot;
    size[tot] = 1, fa[tot] = ch[tot][0] = ch[tot][1] = rev[tot] = 0;
}

bool isRoot(int x){
    return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
}

void push_up(int x){
    size[x] = size[ch[x][0]] + size[ch[x][1]] + 1;
}

void reverse(int x){
    rev[x] ^= 1;
    swap(ch[x][0], ch[x][1]);
}

void push_down(int x){
    if(rev[x]){
        if(ch[x][0]) reverse(ch[x][0]);
        if(ch[x][1]) reverse(ch[x][1]);
        rev[x] ^= 1;
    }
}

void rotate(int x){
    int y = fa[x], z = fa[y], p = (ch[y][1] == x) ^ 1;
    ch[y][p^1] = ch[x][p], fa[ch[x][p]] = y;
    if(!isRoot(y)) ch[z][ch[z][1] == y] = x;
    fa[x] = z, fa[y] = x, ch[x][p] = y;
    push_up(y), push_up(x);
}

void splay(int x){
    int pos = 0; st[++ pos] = x;
    for(int i = x; !isRoot(i); i = fa[i]) st[++ pos] = fa[i];
    while(pos) push_down(st[pos --]);
    while(!isRoot(x)){
        int y = fa[x], z = fa[y];
        if(!isRoot(y)){
            (ch[y][0] == x) ^ (ch[z][0] == y) ? rotate(x) : rotate(y);
        }
        rotate(x);
    }
    push_up(x);
}

void access(int x){
    for(int p = 0; x; p = x, x = fa[x])
        splay(x), ch[x][1] = p, push_up(x);
}

void makeRoot(int x){
    access(x), splay(x), reverse(x);
}

void split(int x, int y){
    makeRoot(x), access(y), splay(y);
}

void link(int x, int y){
    makeRoot(x);
    fa[x] = y;
}

void cut(int x, int y){
    split(x, y);
    fa[x] = ch[y][0] = 0;
    push_up(y);
}

int find(int x, int k){
    int u = x;
    for(int i = 0; k; k >>= 1, i ++)
        if(k & 1) u = p[u][i];
    return u;
}

int main(){

    for(_ = read(); _; _ --){
        tot = 0, full(to, 0);
        n = read();
        t = (int)(log(n) / log(2)) + 1;
        for(int i = 2; i <= n; i ++){
            p[i][0] = read();
        }
        for(int i = 1; i <= t; i ++){
            for(int j = 1; j <= n; j ++){
                p[j][i] = p[p[j][i - 1]][i - 1];
            }
        }
        for(int i = 1; i <= n + 1; i ++) build();
        for(int i = 1; i <= n; i ++){
            to[i] = find(i, read());
            link(i + 1, to[i] + 1);
        }
        m = read();
        while(m --){
            int opt = read();
            if(opt == 1){
                int u = read();
                split(1, u + 1);
                printf("%d
", size[u + 1] - 1);
            }
            else{
                int u = read(), k = read();
                cut(u + 1, to[u] + 1);
                to[u] = find(u, k);
                link(u + 1, to[u] + 1);
            }
        }
    }
    return 0;
}