uva 11922 Permutation Transformer

題目鏈接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=229&page=show_problem&problem=3073

題目描述：給你一個1~n的序列，有m個操作，每次操作就是將下標[a ,b]區間的序列反轉並且將其整體刪除，然後添加到序列尾部

解題思路：這個是劉汝佳白書上的原題，主要使用伸展樹(splay）維護整個區間被反轉的次數。

這也是我第一次寫splay ，寫得過程中出了好多錯誤。splay就是一棵平衡的二叉樹，雖然單次操作複雜度可能會很大，但是可以證明均攤複雜度是O(logn)，好神奇。。。

如果已經學過AVL樹，那就很簡單了，splay的操作和AVL操作很類似，但是編程實現比較簡單，而且有好多小技巧，建議大家自己獨立寫一遍

//#pragma comment(linker,"/STACK:102400000,102400000")
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#define ll long long
#define db double
#define PB push_back
#define lson k<<1
#define rson k<<1|1
using namespace std;

const int N = 100005;

struct node
{
    node *ch[2];
    int v, s;
    bool flip;
    int cmp(int x) const
    {
        int d = x - ch[0]->s;
        if(d == 1) return -1;
        return d <= 0 ? 0 : 1;
    }
    void maintain()
    {
        s = 1 + ch[0]->s + ch[1]->s;
    }
    void pushdown()
    {
        if(flip)
        {
            flip = false;
            swap(ch[0], ch[1]);
            ch[0]->flip = !ch[0]->flip;
            ch[1]->flip = !ch[1]->flip;
        }
    }
};

void rotate(node* &o, int d)
{
    node* k = o->ch[d ^ 1];
    o->ch[d ^ 1] = k->ch[d], k->ch[d] = o;
    o->maintain(), k->maintain();
    o = k;
}

node* root;
node* null = new node();
void splay(node* &o, int k)
{
    o->pushdown();
    int d = o->cmp(k);
    if(d != -1)
    {
        if(d == 1) k -= o->ch[0]->s + 1;
        node* p = o->ch[d];
        p->pushdown();
        int d2 = p->cmp(k);
        if(d2 != -1)
        {
            if(d2==1) k-=p->ch[0]->s+1;
            splay(p->ch[d2], k);
            if(d == d2) rotate(o, d ^ 1);
            else rotate(o->ch[d], d);
        }
        rotate(o, d ^ 1);
    }
}

node* merge(node* left, node* right)
{
    splay(left, left->s);
    left->ch[1] = right;
    left->maintain();
    return left;
}

void split(node* o, int k, node* &left, node* &right)
{
    splay(o, k);
    left = o;
    right = o->ch[1];
    o->ch[1] = null;
    left->maintain();
}

node seq[N];
node* build(int l, int r,int &n)
{
    if(l > r) return null;
    if(l == r)
    {
        node *tmp = &seq[++n];
        tmp->s = 1;
        tmp->v = n-1;
        tmp->ch[0] = tmp->ch[1] = null;
        tmp->flip = false;
        return tmp;
    }
    int mid = (l + r) >> 1;
    node *L=build(l,mid-1,n);
    node *tmp=&seq[++n];
    tmp->ch[0] = L;
    tmp->flip = false;
    tmp->v = n-1;
    tmp->ch[1] = build(mid + 1, r,n);
    tmp->maintain();
    return tmp;
}
void init(int n)
{
    null->s = 0;
    int num=0;
    root = build(1, n + 1,num);
}

void print(node *o)
{
    if(o == null) return;
    o->pushdown();
    print(o->ch[0]);
    if(o->v > 0) printf("%d\n", o->v);
    print(o->ch[1]);
}

int main()
{
#ifdef PKWV
    freopen("in.in", "r", stdin);
#endif // PKWV
    int n, m;
    scanf("%d%d", &n, &m);
    init(n);
    while(m--)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        node *left, *right;
        node *l, *r;
        split(root, a, left, right);
        split(right, b - a + 1, l, r);
        l->flip ^= 1;
        root = merge(merge(left, r), l);
    }
    print(root);
    return 0;
}