【splay tree】 UVA 11922 Permutation Transformer

题目中只有一种操作，把一段区间翻转加到序列后面。。伸展树虽然在区间合并和区间会破坏序列的绝对关系，但是在各种操作中序列的中序遍历是不变的。。。因此我们可以维护一颗伸展树。。然后打上lazy标记，最后求一下最终序列就行了。。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>  
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 805
#define maxm 400005
#define eps 1e-10
#define mod 1000000009
#define INF 99999999  
#define lowbit(x) (x&(-x))  
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;

struct node
{
	int s, v, flip;
	node *ch[2];
	int cmp(int x)
	{
		if(x == ch[0]->s+1) return -1;
		if(x > ch[0]->s+1) return 1;
		else return 0;
	}
	void maintain(void)
	{
		s = ch[0]->s + ch[1] ->s + 1;
	}
	void pushdown(void)
	{
		if(!flip) return;
		swap(ch[0], ch[1]);
		ch[0]->flip^=1;
		ch[1]->flip^=1;
		flip = 0;
	}
}*null, *root;
int n, m;

void init(int x)
{
	null = new node;
	null->s = null->v = null->flip = 0;
	null->ch[0] = null->ch[1] = NULL;
	root = new node;
	root->v = root->flip = 0, root->ch[0] = root->ch[1] = null;
	root->maintain();
	node *p;
	for(int i = 1; i <= n; i++) {
		p = new node;
		p->v = i, p->flip = 0;
		p->ch[0] = root, p->ch[1] = null;
		root = p, root->maintain();
	}
}
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;
}
void splay(node* &o, int k)
{
	o->pushdown();
	int d = o->cmp(k);
	if(d == 1) k -= o->ch[0]->s + 1;
	if(d != -1) {
		node *p = o->ch[d];
		p->pushdown();
		int d2 = p->cmp(k);
		int k2 = (d2 == 0 ? k : k - p->ch[0]->s - 1);
		if(d2 != -1) {
			splay(p->ch[d2], k2);
			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);
	right = o->ch[1];
	o->ch[1] = null;
	left = o;
	left->maintain();
}
void solve(node* &o)
{
	o->pushdown();
	if(o->ch[0] != null) solve(o->ch[0]);
	if(o->v != 0) printf("%d\n", o->v);
	if(o->ch[1] != null) solve(o->ch[1]);
}
void work(void)
{
	int a, b;
	while(m--) {
		scanf("%d%d", &a, &b);
		node *left, *right, *mid, *o;
		split(root, a, left, o);
		split(o, b-a+1, mid, right);
		mid->flip^=1;
		root = merge(merge(left, right), mid);
	}
}
int main(void)
{
	while(scanf("%d%d", &n, &m)!=EOF) {
		init(n);
		work();
		solve(root);
	}
	return 0;
}