Learning - I played with a sort Splay

Splay learning - I Splay fight sorting

Description

Given the number N, your task is to put the number of small to large order n ^ _ ^

Input

The first line of an integer, N (1 <= n < = 200000), the number N expressed
Then there are N rows, each row a number

Output

N output lines, N is the number of already sorted

Sample Input

5
2
3
1
4
5

Sample Output

1
2
3
4
5

The problem I believe we will use the sorting algorithm, the reason why I use Splay, because they can not on OJ splay version of the title. . .

As a result, I can only try to insert, Splay operation

Code

//请自行忽略这两行注释
//#pragma GCC optimize(2)
//#pragma G++ optimize(2)
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int a[200010];
struct node{
    int son[2],v,t,f,size;
    node(){t=f=son[0]=son[1]=0;size=1;}
};
node no[200010];int cntn=1,root=1;
int getson(int x){return (int)(no[no[x].f].son[1]==x);}
void updata(int x)
{
    no[x].size=no[no[x].son[0]].size+no[no[x].son[1]].size+1;
}
void rotate(int x)
{
    if(!no[x].f) return ;
    int dirx,dirfa,fa,gf;
    fa=no[x].f, gf=no[fa].f;
    dirx=getson(x),dirfa=getson(fa);
    no[fa].son[dirx]=no[x].son[dirx^1];
    no[no[fa].son[dirx]].f=fa;
    no[x].son[dirx^1]=fa;
    if(gf>0) no[gf].son[dirfa]=x;
    no[fa].f=x,no[x].f=gf;
    updata(fa);updata(x);
}
void splay(int x,int y)
{
    int z;
    while(no[x].f!=y) rotate(x);
    updata(x);
    if(y==0) root=x;
}
void ins(int v)
{
    int x=root;
    while(x)
    {
        if(v==no[x].v){no[x].t++;break;}
        if(v<no[x].v)
        {
            if(!no[x].son[0]) ++cntn,no[x].son[0]=cntn,no[cntn].v=v,no[cntn].f=x;
            x=no[x].son[0];
        }
        else
        {
            if(!no[x].son[1]) ++cntn,no[x].son[1]=cntn,no[cntn].v=v,no[cntn].f=x;
            x=no[x].son[1];
        }
    }
    splay(x,0);
}
void DFS(int x)
{
    if(x==0) return ;
    DFS(no[x].son[0]);
    for(int i=1;i<=no[x].t;i++) printf("%d\n",no[x].v);
    DFS(no[x].son[1]);
}
int main()
{
    freopen("sort.in","r",stdin);
    freopen("sort.out","w",stdout);
    int n,i,u,v,t;
    scanf("%d",&n);
    for(i=1;i<=n;i++) scanf("%d",&a[i]);
    no[root].v=a[1],no[root].t=1;
    for(i=2;i<=n;i++)
    {
        ins(a[i]);
    }
    DFS(root);
    return 0;
}

Submitted after the OJ. . .

TLE0！？

Me: "It must be splay constant too."

Then opened the O2

Not yet? Oh, ambidextrous

void splay(int x,int y)
{
    int z;
    while(no[x].f!=y)
    {
        z=no[x].f;
        if(no[z].f!=y&&getson(z)==getson(x)) rotate(z);
        else rotate(x);
    }
    updata(x);
    if(y==0) root=x;
}

I: emmm ...

Then being passed with a random exchange Gou

probably after a few days,

a small room appeared similar sound

A student: I Splay TLE up!

So I baidu a bit,
splay timeout

Splay的伸展

当我们把某一个点x一路旋转至某个位置时……我们可以考虑一次性进行两次旋转来让伸展之后的树更平衡。
我们一般进行如下的分类讨论：
若x的父节点即为伸展目标：直接旋转x即可。
若x的父节点不是伸展目标，且x、x的父亲、x的父亲的父亲三点共线
（即x是x的父亲的左儿子，x的父亲的父亲也是x的父亲的左儿子； 或x是x的父亲的右儿子，x的父亲的父亲也是x的父亲的右儿子）。

这时先旋转x的父亲，再旋转x。

若三点不共线，对x进行两次旋转即可。

额。。。

void splay(int x,int y)
{
    int z;
    while(no[x].f!=y)
    {
        z=no[x].f;
        if(no[z].f!=y&&getson(z)==getson(x)) rotate(z);
        //else rotate(x);
        rotate(x);
    }
    updata(x);
    if(y==0) root=x;
}

完整代码

//#pragma GCC optimize(2)
//#pragma G++ optimize(2)
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int a[200010];
struct node{
    int son[2],v,t,f,size;
    node(){t=f=son[0]=son[1]=0;size=1;}
};
node no[200010];int cntn=1,root=1;
int getson(int x){return (int)(no[no[x].f].son[1]==x);}
void updata(int x)
{
    no[x].size=no[no[x].son[0]].size+no[no[x].son[1]].size+1;
}
void rotate(int x)
{
    if(!no[x].f) return ;
    int dirx,dirfa,fa,gf;
    fa=no[x].f, gf=no[fa].f;
    dirx=getson(x),dirfa=getson(fa);
    no[fa].son[dirx]=no[x].son[dirx^1];
    no[no[fa].son[dirx]].f=fa;
    no[x].son[dirx^1]=fa;
    if(gf>0) no[gf].son[dirfa]=x;
    no[fa].f=x,no[x].f=gf;
    updata(fa);updata(x);
}
void splay(int x,int y)
{
    int z;
    while(no[x].f!=y)
    {
        z=no[x].f;
        if(no[z].f!=y&&getson(z)==getson(x)) rotate(z);
        rotate(x);
    }
    updata(x);
    if(y==0) root=x;
}
void ins(int v)
{
    int x=root;
    while(x)
    {
        if(v==no[x].v){no[x].t++;break;}
        if(v<no[x].v)
        {
            if(!no[x].son[0]) ++cntn,no[x].son[0]=cntn,no[cntn].v=v,no[cntn].f=x;
            x=no[x].son[0];
        }
        else
        {
            if(!no[x].son[1]) ++cntn,no[x].son[1]=cntn,no[cntn].v=v,no[cntn].f=x;
            x=no[x].son[1];
        }
    }
//  Db();
    splay(x,0);
//  Db();
}
void DFS(int x)
{
    if(x==0) return ;
    DFS(no[x].son[0]);
    for(int i=1;i<=no[x].t;i++) printf("%d\n",no[x].v);
    DFS(no[x].son[1]);
}
int main()
{
    freopen("sort.in","r",stdin);
    freopen("sort.out","w",stdout);
    int n,i,u,v,t;
    scanf("%d",&n);
    srand(19260817);
    for(i=1;i<=n;i++) scanf("%d",&a[i]);
    no[root].v=a[1],no[root].t=1;
    for(i=2;i<=n;i++)
    {
        ins(a[i]);
    }
    DFS(root);
    return 0;
}