bzoj3223: Tyvj 1729 文艺平衡树 splay裸题

splay区间翻转即可

/**************************************************************
    Problem: 3223
    User: walfy
    Language: C++
    Result: Accepted
    Time:2304 ms
    Memory:5444 kb
****************************************************************/
 
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)
 
using namespace std;
 
const double eps=1e-6;
const int N=2000+10,maxn=5000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;
 
struct Splay{
    struct Node{
        Node* ch[2];
        int v;
        int s;
        int 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)//类似于线段树的lazy标记
            {
                flip=0;
                swap(ch[0],ch[1]);
                ch[0]->flip = !(ch[0]->flip);
                ch[1]->flip = !(ch[1]->flip);
            }
        }
    };
    Node* 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 splay(Node* &o,int k)
    {
        o->pushdown();
        int d = o->cmp(k);
        if(d==1)k -= o->ch[0]->s + 1;//利用二叉树性质
        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);//把排名最大的数splay到根
        left->ch[1] = right;
        left->maintain();
        return left;
    }
    void split(Node* o,int k,Node* &left,Node* &right)
    {
        splay(o,k);//把排名为k的节点splay到根，右侧子树所有节点排名比k大，左侧小
        right = o->ch[1];
        o->ch[1] = null;
        left = o;
        left->maintain();
    }
    Node *root,*left,*right;
    void init(int sz)
    {
        null=new Node;
        null->s=0;
        root=new Node;
        root->v=1;root->flip=0;
        root->ch[0]=root->ch[1]=null;
        root->maintain();
        Node* p;
        for(int i=2;i<=sz;i++)
        {
            p=new Node;
            p->v=i;p->s=p->flip=0;
            p->ch[0]=root,p->ch[1]=null;
            root=p;
            root->maintain();
        }
    }
    void gao(int n,int l,int r)
    {
        if(l==1&&r==n)root->flip^=1;
        else if(l==1)
        {
            split(root,r,left,right);
            left->flip^=1;
            root=Merge(left,right);
        }
        else if(r==n)
        {
            split(root,l-1,left,right);
            right->flip^=1;
            root=Merge(left,right);
        }
        else
        {
            Node *mid;
            split(root,r,left,right);
            split(left,l-1,left,mid);
            mid->flip^=1;
            root=Merge(Merge(left,mid),right);
        }
    }
    void print(Node *o)
    {
        o->pushdown();
        if(o->ch[0]!=null)print(o->ch[0]);
        printf("%d ",o->v);
        if(o->ch[1]!=null)print(o->ch[1]);
    }
}sp;
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    sp.init(n);
    for(int i=0;i<m;i++)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        sp.gao(n,l,r);
    }
    sp.print(sp.root);
    return 0;
}
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