BZOJ_P3223&Tyvj_P1729 文艺平衡树(SplayTree区间翻转)

传送门
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 2445 Solved: 1368
[Submit][Status][Discuss]
Description

您需要写一种数据结构（可参考题目标题），来维护一个有序数列，其中需要提供以下操作：翻转一个区间，例如原有序序列是5 4 3 2 1，翻转区间是[2,4]的话，结果是5 2 3 4 1

Input
第一行为n,m n表示初始序列有n个数，这个序列依次是(1,2……n-1,n) m表示翻转操作次数
接下来m行每行两个数[l,r] 数据保证 1<=l<=r<=n

Output
输出一行n个数字，表示原始序列经过m次变换后的结果

Sample Input
5 3
1 3
1 3
1 4

Sample Output
4 3 2 1 5

HINT
N,M<=100000

Source
平衡树

#include<cstdio>#include<cstdlib>#include<climits>#include<iostream>using namespace std;struct Node{    Node *ch[2];    int r,v,s;int b;    void pushdown(){if(b){b=0;swap(ch[0],ch[1]);ch[0]->b^=1;ch[1]->b^=1;}}    Node(int v,Node *nl):v(v){r=rand();b=0;ch[0]=ch[1]=nl;s=1;}    void maintain(){s=1;s+=ch[0]->s;s+=ch[1]->s;}}*root,*null;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 insert(Node* &o,int x){    if(o==null){o=new Node(x,null);return;}    insert(o->ch[1],x);    if(o->ch[1]->r>o->r) rotate(o,0);    else o->maintain();}int cmprk(Node* o,int k){    if(o->ch[0]->s+1==k) return -1;    if(o->ch[0]->s>=k) return 0;else return 1;}void splay(Node* &o,int k){    if(o==null) return;    o->pushdown();    int d=cmprk(o,k);    if(d==1) k-=o->ch[0]->s+1;    if(d!=-1&&o->ch[d]!=null){        Node *p=o->ch[d];p->pushdown();        int d2=cmprk(p,k);        if(d2!=-1&&p->ch[d2]!=null){            int k2=(d2==0?k:k-p->ch[0]->s-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){//合并      splay(left,left->s);left->ch[1]=right;    left->maintain();return left;}void split(Node *o,int k,Node* &left,Node* &right){//分裂 前K小在left中其余在right中     if(k==0){left=null,right=o;return;}    if(k==o->s){left=o,right=null;return;}    splay(o,k);left=o;right=o->ch[1];    o->ch[1]=null;left->maintain(); }void init(){    null=new Node(0,0);null->r=INT_MAX;null->s=null->v=0;    null->ch[0]=null->ch[1]=null;root=null;}void print(Node *o){    if(o==null) return;    o->pushdown();    print(o->ch[0]);    if(o->r!=-888) printf("%d ",o->v);    print(o->ch[1]);}int n,m,x,l,r;Node *ll,*tmp,*mm,*rr;int main(){    scanf("%d%d",&n,&m);init();    insert(root,0);root->r=-888;    for(int i=1;i<=n;i++) insert(root,i);//  print(root);putchar('\n');    while(m--){        scanf("%d%d",&l,&r);        split(root,l,ll,rr);        split(rr,r-l+1,mm,rr);        mm->b^=1;        root=merge(merge(ll,mm),rr);//      print(root);putchar('\n');    }    print(root);    return 0;}