【BZOJ3233】【tyvj1729】文艺平衡树

原题传送门

解题思路：裸平衡树操作，支持区间翻转即可，这里写了无旋treap。

其实平衡树的区间操作就和线段树差不多，你用个标记搞一下就好了，，，，，

#include <stdio.h>
#define r register
#define getchar() (S==TT&&(TT=(S=BB)+fread(BB,1,1<<15,stdin),S==TT)?EOF:*S++)
char BB[1<<15],*TT=BB,*S=BB;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9') f=ch=='-'?-1:1,ch=getchar();
    while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
    return x*f;
}
namespace Treap{
    inline int Rand(){
        static int x=23333;
        return x^=x<<13,x^=x>>17,x^=x<<5;
    }
    struct node{
        node *ls,*rs;
        int val,sz,pri;
        bool rev;
        node(int x):val(x),rev(0),sz(1),pri(Rand()){ls=rs=NULL;}
        void combine(){sz=(ls?ls->sz:0)+(rs?rs->sz:0)+1;}
    }*root;
    struct Droot{node *a,*b;};
    inline int Size(node *x){return x?x->sz:0;}
    inline void swap(node *&x,node *&y){r node *t=x;x=y;y=t;}
    inline node *reverse(node *x){if (!x) return NULL;swap(x->ls,x->rs);x->rev^=1;return x;}
    inline void pushdown(node *x){if (x->rev) reverse(x->ls),reverse(x->rs),x->rev=0;}
    node *merge(node*a,node*b){
        if (!a) return b;if (!b) return a;
        if (a->pri<b->pri){
            pushdown(a);a->rs=merge(a->rs,b);
            a->combine();return a;
        }else{
            pushdown(b);b->ls=merge(a,b->ls);
            b->combine();return b;
        }
    }
    Droot split(node *x,int k){
        if (x==NULL) return (Droot){NULL,NULL};
        r Droot y;pushdown(x);
        if (k<=Size(x->ls)){
            y=split(x->ls,k);x->ls=y.b;
            x->combine();y.b=x;
        }else{
            y=split(x->rs,k-Size(x->ls)-1);
            x->rs=y.a;x->combine();y.a=x;
        }return y;
    }
    inline void Reverse(int L,int R){
        r Droot x=split(root,L-1);
        r Droot y=split(x.b,R-L+1);
        root=merge(merge(x.a,reverse(y.a)),y.b);
    }
}using namespace Treap;
int n,q;
void init(){
    n=read(),q=read();
    for (int i=1; i<=n; ++i) root=merge(root,new node(i));
}
void solve(){
    while(q--){
        r int L=read(),R=read();
        Reverse(L,R);
    }for (r int i=1; i<=n; ++i) {
        r Droot y=split(root,1);
        printf("%d ",y.a->val); 
        root=y.b;
    }
}
int main(){init(); solve(); return 0;}