题解:
每次reverse(l,r)
把l-1转到根,r+1变成他的右儿子,给r+1的左儿子打个标记就是一次反转操作了
每次find和dfs输出的时候下放标记,把左儿子和右儿子换一下
记得建树的时候建0到n+1
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 100010
#define which(x) (ls[fa[(x)]]==(x))
using namespace std;
int sz[N],ls[N],rs[N],fa[N],n,m,root,flag[N],id[N],idx;
void upt(int x) {sz[x]=+sz[ls[x]]+sz[rs[x]];}
int read()
{
int ret=,neg=;
char j=getchar();
for (;j>'' || j<'';j=getchar())
if (j=='-') neg=-;
for (;j>='' && j<='';j=getchar())
ret=ret*+j-'';
return ret*neg;
}
void write(int x)
{
if (x<) x=-x;
if (x>=) write(x/);
putchar(x%+'');
}
void swap_son(int u)
{
if (!u) return;
flag[u]^=;
swap(ls[u],rs[u]);
upt(u);
}
void pushdown(int u)
{
if (!flag[u]) return ;
swap_son(ls[u]),swap_son(rs[u]);
flag[u]=;
}
int Build(int l, int r)
{
int mid=l+r>>,u=++idx;
id[u]=mid;
if(mid>l) ls[u]=Build(l,mid-),fa[ls[u]]=u;
if(mid<r) rs[u]=Build(mid+,r),fa[rs[u]]=u;
upt(u);
return u;
}
void Rotate(int u)
{
int v=fa[u],w=fa[v],b=which(u)?rs[u]:ls[u];
if(w) which(v)?ls[w]=u:rs[w]=u;
which(u)?(ls[v]=b,rs[u]=v):(rs[v]=b,ls[u]=v);
fa[u]=w,fa[v]=u;
if(b) fa[b]=v;
upt(v),upt(u);
}
void Splay(int u, int tar)
{
while(fa[u] != tar)
{
if(fa[fa[u]] != tar)
{
if(which(u) == which(fa[u])) Rotate(fa[u]);
else Rotate(u);
}
Rotate(u);
}
if(!tar) root = u;
}
int find(int k)
{
int u=root;
pushdown(u);
while (sz[ls[u]]+!=k && u)
{
if (sz[ls[u]]>=k) u=ls[u];
else k-=sz[ls[u]]+,u=rs[u];
pushdown(u);
}
return u;
}
void rev(int l,int r)
{
int u=find(l-),v=find(r+);
Splay(u,);
Splay(v,u);
swap_son(ls[rs[root]]);
}
void dfs(int u)
{
pushdown(u);
if (ls[u]) dfs(ls[u]);
if (id[u]> && id[u]<n+) write(id[u]-),putchar(' ');
if (rs[u]) dfs(rs[u]);
}
int main()
{
scanf("%d%d",&n,&m);
root=Build(,n+);
for (int i=,l,r;i<=m;i++)
l=read(),r=read(),rev(l+,r+);
dfs(root);
return ;
}