BZOJ 2002 弹飞绵羊

LCT

刚学LCT,对LCT的性质不太熟练,还需要多多练习。。

对每一个点,将其与它能够到达的点连一条虚边。弹出去的话就用n+1这个节点表示。
第一种操作我们需要从LCT的性质入手,问的问题其实就是x通过多少条边可以到达n+1这个点。。那么我们可以把他们两拉成一条链(也就是split(n + 1, x)),这样就把x splay到根了,根据LCT的性质,在x和n+1联通的这棵splay中,x一定没有右子树,因为他是深度最大的点。那么左子树所有点就是原树中x到n+1的路径上的所有点了。。直接输出答案size就好了
第二种操作很简单断边再连边就行了,因为这题的边一定合法,所以没什么需要考虑的特殊情况

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
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 a % b ? gcd(b, a % b) : b; }
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 = 200005;
int n, tot, ch[N][2], size[N], fa[N], st[N], rev[N], f[N];

int newNode(){
    size[++tot] = 1, fa[tot] = ch[tot][0] = ch[tot][1] = 0;
    return tot;
}

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

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

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

void push_down(int x){
    if(rev[x]){
        reverse(ch[x][0]), 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)){
            if((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 link(int x, int y){
    makeRoot(x), fa[x] = y, push_up(y);
}

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

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

int main(){

    n = read();
    for(int i = 1; i <= n + 1; i ++) newNode();
    for(int i = 1; i <= n; i ++){
        int k = read(), t = min(i + k, n + 1);
        link(i, t), f[i] = t;
    }
    int m = read();
    while(m --){
        int opt = read();
        if(opt == 1){
            int x = read(); x ++;
            split(n + 1, x); printf("%d\n", size[x] - 1);
        }
        else{
            int x = read(), y = read(); x ++;
            cut(x, f[x]);
            int t = min(x + y, n + 1);
            link(x, t), f[x] = t;
        }
    }
    return 0;
}
posted @ 2019-04-12 17:08  清楚少女ひなこ  阅读(68)  评论(0编辑  收藏