此为平衡树系列第二道：文艺平衡树您需要写一种数据结构，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
输入
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
输出
输出一行n个数字，表示原始序列经过m次变换后的结果
样例输入
5 3
1 3
1 3
1 4
样例输出
4 3 2 1 5
提示
n,m<=100000
非旋转treap相比于旋转treap支持区间操作，所以对于这道题只需要每次把树拆成[1~l-1]、[l~r]、[r+1~tot]三段，然后把[l~r]打上标记进行翻转，再把三段区间合并就行了。对于打标记的点只有在查到这个点时才翻转，如果一个点被打了两次标记就相当于不翻转。
最后附上代码.
指针版
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
char *p1,*p2,buf[100000];
#define nc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int rd() {int x=0,f=1; char c=nc(); while(!isdigit(c)) {if(c=='-') f=-1; c=nc();} while(isdigit(c)) x=(((x<<2)+x)<<1)+(c^48),c=nc(); return x*f;}
ll rd2() {ll x=0,f=1; char c=nc(); while(!isdigit(c)) {if(c=='-') f=-1; c=nc();} while(isdigit(c)) x=(((x<<2)+x)<<1)+(c^48),c=nc(); return x*f;}
int n,m;
int L,R;
struct treap
{int size;int rnd;int val;int rev;treap *ls,*rs;treap(){}treap(int k,treap *son){size=1;rnd=rand();val=k;ls=rs=son;rev=0;}void pushup(){size=ls->size+rs->size+1;}void pushdown(){if(rev){rev^=1;ls->rev^=1;rs->rev^=1;swap(ls,rs);}}
}tr[100010],nil;
typedef treap* node;
node root,cnt,null;
void init()
{nil=treap(0,NULL);null=&nil;null->ls=null->rs=null;null->size=0;root=null;cnt=tr;
}
node build(int x)
{*cnt=treap(x,null);return cnt++;
}
node merge(node x,node y)
{if(x==null){return y; }if(y==null){return x;}x->pushdown();y->pushdown();if(x->rnd<y->rnd){x->rs=merge(x->rs,y);x->pushup();return x;}else{y->ls=merge(x,y->ls);y->pushup();return y;}
}
void split(node rt,node &x,node &y,int k)
{if(rt==null){x=y=null;return ;}rt->pushdown();if(rt->ls->size>=k){y=rt;split(rt->ls,x,y->ls,k);rt->pushup();}else{x=rt;split(rt->rs,x->rs,y,k-rt->ls->size-1);rt->pushup();}
}
void print(node rt)
{rt->pushdown();if(rt->ls!=null){print(rt->ls);}printf("%d",rt->val);if(--n!=0){printf(" ");}if(rt->rs!=null){print(rt->rs);}
}
node build_tree(int l,int r)
{if(l==r){return build(l);}int mid=(l+r)>>1;return merge(build_tree(l,mid),build_tree(mid+1,r));
}
int main()
{init();n=rd();m=rd();root=build_tree(1,n);while(m--){L=rd();R=rd();node x,y,z;split(root,y,z,R);split(y,x,y,L-1);y->rev^=1;root=merge(merge(x,y),z);}print(root);
}
非指针版
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstring>
using namespace std;
int n,m;
int r[100010];
int v[100010];
int s[100010][3];
int size[100010];
bool d[100010];
int root;
int tot;
int L,R;
int x,y,z;
int flag;
void pushup(int x)
{size[x]=size[s[x][0]]+size[s[x][1]]+1;
}
int build(int x)
{v[++tot]=x;r[tot]=rand();size[tot]=1;return tot;
}
void pushdown(int x)
{swap(s[x][0],s[x][1]);if(s[x][0]){d[s[x][0]]^=1;}if(s[x][1]){d[s[x][1]]^=1;}d[x]=0;
}
int merge(int x,int y)
{if(!x||!y){return x+y;}if(r[x]<r[y]){if(d[x]){pushdown(x);}s[x][1]=merge(s[x][1],y);pushup(x);return x;}else{if(d[y]){pushdown(y);}s[y][0]=merge(x,s[y][0]);pushup(y);return y;}
}
void split(int i,int k,int &x,int &y)
{if(!i){x=y=0;return ;}if(d[i]){pushdown(i);}if(size[s[i][0]]<k){x=i;split(s[i][1],k-size[s[i][0]]-1,s[i][1],y);}else{y=i;split(s[i][0],k,x,s[i][0]);}pushup(i);
}
void print(int x)
{if(!x){return ;}if(d[x]){pushdown(x);}print(s[x][0]);if(flag==0){printf("%d",v[x]);flag=1;}else{printf(" %d",v[x]);}print(s[x][1]);
}
int main()
{srand(12378);scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){root=merge(root,build(i));}for(int i=1;i<=m;i++){scanf("%d%d",&L,&R);split(root,L-1,x,y);split(y,R-L+1,y,z);d[y]^=1;root=merge(merge(x,y),z);}print(root);return 0;
}
