uva 11922 Permutation Transformer

来源：互联网发布：色环电阻计算器软件编辑：程序博客网时间：2024/04/28 08:06

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=229&page=show_problem&problem=3073

题目描述：给你一个1~n的序列，有m个操作，每次操作就是将下标[a ,b]区间的序列反转并且将其整体删除，然后添加到序列尾部

解题思路：这个是刘汝佳白书上的原题，主要使用伸展树(splay）维护整个区间被反转的次数。

这也是我第一次写splay ，写得过程中出了好多错误。splay就是一棵平衡的二叉树，虽然单次操作复杂度可能会很大，但是可以证明均摊复杂度是O(logn)，好神奇。。。

如果已经学过AVL树，那就很简单了，splay的操作和AVL操作很类似，但是编程实现比较简单，而且有好多小技巧，建议大家自己独立写一遍

//#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|1using 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;}

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