題目:http://www.lydsy.com/JudgeOnline/problem.php?id=3223
帶翻轉的splay 隻需要把l-1提到根位置,r+1提到根的右子樹,那麼r+1的左子樹就是[l,r]
對反轉次數打标記
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define lson i<<1
#define rson i<<1|1
using namespace std;
const int N=1e5+5;
const int inf=0x3f3f3f3f;
int n,m;
int f[N],ch[N][2],mark[N],sz[N],key[N];
int root,tot;
int a[N];
inline int get(int x)
{
return ch[f[x]][1]==x;
}
inline void pushup(int x)
{
sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+1;
}
void pushdown(int x)
{
if (x&&mark[x])
{
mark[ch[x][0]]^=1;
mark[ch[x][1]]^=1;
swap(ch[x][0],ch[x][1]);
mark[x]=0;
}
}
int build(int l,int r,int fa)
{
if (l>r) return 0;
int mid=(l+r)>>1;
int now=++tot;
f[now]=fa;key[now]=a[mid];mark[now]=0;
ch[now][0]=build(l,mid-1,now);
ch[now][1]=build(mid+1,r,now);
pushup(now);
return now;
}
int Find(int x)
{
int now=root;
while(1)
{
pushdown(now);
if (x<=sz[ch[now][0]])
now=ch[now][0];
else
{
x=x-sz[ch[now][0]]-1;
if (!x) return now;
now=ch[now][1];
}
}
}
void Rotate(int x)
{
pushdown(f[x]);
pushdown(x);
int fa=f[x],ff=f[fa],kind=get(x);
ch[fa][kind]=ch[x][kind^1];f[ch[x][kind^1]]=fa;
ch[x][kind^1]=fa;f[fa]=x;
f[x]=ff;
if (ff)
ch[ff][ch[ff][1]==fa]=x;
pushup(fa);
pushup(x);
}
void splay(int x,int y)
{
for(int fa;(fa=f[x])!=y;Rotate(x))
if (f[fa]!=y)
Rotate((get(x)==get(fa))?fa:x);
if (y==0) root=x;
}
void out(int now)
{
pushdown(now);
if (ch[now][0]) out(ch[now][0]);
if (key[now]!=inf&&key[now]!=-inf)
printf("%d ",key[now]);
if (ch[now][1]) out(ch[now][1]);
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(sz,0,sizeof(sz));
tot=0;
a[1]=-inf;a[n+2]=inf;
for(int i=1;i<=n;i++)
a[i+1]=i;
root=build(1,n+2,0);
int x,y;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
// x-1 y+1 還有-inf 是以就是x-1+1 y+1+1
int xx=Find(x);
int yy=Find(y+2);
splay(xx,0);
splay(yy,xx);
mark[ch[ch[root][1]][0]]^=1;
}
out(root);
printf("\n");
}
return 0;
}
轉載于:https://www.cnblogs.com/bk-201/p/7390148.html