P3203 [HNOI2010]弹飞绵羊(LCT)

弹飞绵羊

题目传送门

解题思路

LCT。

将每个节点的权值设为\(1\)，连接\(i\)和\(i+ki\)，被弹飞就连上\(n\)，维护权值和\(sum[]\)。从\(j\)弹飞需要的次数就是\(split(j,n)\)后，\(sum[i]-1\)的值。修改弹力系数，即为断开\(i\)和旧的\(i+ki\)的连接，然后连上\(i\)和新的\(i+ki\)。

为了方便，以下代码把下标都加一了，即原编号变为\(1-n\)，弹飞设为\(n+1\)。

代码如下

#include <bits/stdc++.h>

using namespace std;

const int N = 200005;

int fa[N], ch[N][2], siz[N], rev[N], sta[N], v[N];

inline bool get(int x)
{
    return ch[fa[x]][1] == x;
}

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

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

inline void rotate(int x)
{
    int y = fa[x], z = fa[y];
    bool u = get(x);
    ch[y][u] = ch[x][u^1], fa[ch[x][u^1]] = y;
    if(!isroot(y))
        ch[z][get(y)] = x;
    fa[x] = z;
    ch[x][u^1] = y, fa[y] = x;
    push_up(y), push_up(x);
}

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

inline void push_down(int x)
{
    if(rev[x]){
        pushr(ch[x][0]), pushr(ch[x][1]);
        rev[x] = 0;
    }
}

inline void splay(int x)
{
    int pos = 0;
    sta[++pos] = x;
    for(int i = x; !isroot(i); i = fa[i])
        sta[++pos] = fa[i];
    while(pos)
        push_down(sta[pos--]);
    while(!isroot(x)){
        int y = fa[x];
        if(!isroot(y))
            get(x) == get(y) ? rotate(y): rotate(x);
        rotate(x);
    }
}

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

inline void make_root(int x)
{
    access(x); splay(x);
    pushr(x);
}

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

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

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

int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++){
        scanf("%d", &v[i]);
        siz[i] = 1;
    }
    siz[n + 1] = 1;
    for(int i = 1; i <= n; i ++){
        if(i + v[i] <= n)
            link(i, i + v[i]);
        else
            link(i, n + 1);
    }
    int m;
    scanf("%d", &m);
    for(int i = 1; i <= m; i ++){
        int opt, x;
        scanf("%d%d", &opt, &x);
        ++x;
        if(opt == 1){
            split(n + 1, x);
            printf("%d\n", siz[x] - 1);
        }
        else {
            int k;
            scanf("%d", &k);
            if(x + v[x] <= n)
                cut(x, x + v[x]);
            else
                cut(x, n + 1);
            if(x + k <= n)
                link(x, x + k);
            else
                link(x, n + 1);
            v[x] = k;
        }
    }
    return 0;
}