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【BZOJ-1552&3506】robotic sort&排序机械臂 Splay

1552: [Cerc2007]robotic sort

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 806  Solved: 329
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Description

技术分享

Input

输入共两行,第一行为一个整数N,N表示物品的个数,1<=N<=100000。第二行为N个用空格隔开的正整数,表示N个物品最初排列的编号。

Output

输出共一行,N个用空格隔开的正整数P1,P2,P3…Pn,Pi表示第i次操作前第i小的物品所在的位置。 注意:如果第i次操作前,第i小的物品己经在正确的位置Pi上,我们将区间[Pi,Pi]反转(单个物品)。

Sample Input

6
3 4 5 1 6 2

Sample Output

4 6 4 5 6 6

HINT

Source

HNOI2009集训Day6

Solution

Splay基本操作...

我们首先对给出的序列排序,第一关键字是编号,第二关键字是时间戳,然后用splay去维护这个有顺序的序列就可以了

或者可以考虑维护min和pos

Splay要常写!调的跟*一样慢

Code

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;inline int read(){    int x=0; char ch=getchar();    while (ch<0 || ch>9) {ch=getchar();}    while (ch>=0 && ch<=9) {x=x*10+ch-0; ch=getchar();}    return x;}#define MAXN 100010int N,pos[MAXN];struct Node{int id,t;}a[MAXN];inline bool cmp(Node A,Node B) {return A.id==B.id? A.t<B.t:A.id<B.id;}namespace SplayTree{    int size[MAXN],root,sz,fa[MAXN],son[MAXN][2]; bool rev[MAXN];    #define ls(x) son[x][0]    #define rs(x) son[x][1]    inline void Update(int now) {size[now]=size[ls(now)]+size[rs(now)]+1;}    inline void PushDown(int now)    {        if (!rev[now]) return;        rev[ls(now)]^=1; rev[rs(now)]^=1; swap(ls(now),rs(now)); rev[now]=0;    }    inline bool Right(int now) {return son[fa[now]][1]==now;}    inline void rotate(int now)        {        PushDown(fa[now]); PushDown(now);        int f=fa[now],gf=fa[f],wh=Right(now);        son[f][wh]=son[now][wh^1]; fa[son[f][wh]]=f;        fa[f]=now; son[now][wh^1]=f; fa[now]=gf;        if (gf) son[gf][son[gf][1]==f]=now;        Update(f); Update(now);    }    inline void splay(int now,int tar)    {        for (int f; (f=fa[now])!=tar; rotate(now))            if (fa[f]!=tar) rotate(Right(now)==Right(f)? f:now);        if (!tar) root=now;        }    inline int GetX(int k)    {        int x=root;        while (1)            {                PushDown(x);                if (k<=size[ls(x)]) x=ls(x);                    else if (k==size[ls(x)]+1) return x;                        else k-=size[ls(x)]+1,x=rs(x);            }        return -1;    }    inline int GetK(int x) {splay(x,0); return size[ls(x)];}    inline int BuildTree(int l,int r,int last)    {        if (r<l) return 0;        int mid=(l+r)>>1,now=++sz;        pos[a[mid].id]=now; fa[now]=last;        int lson=BuildTree(l,mid-1,now),rson=BuildTree(mid+1,r,now);        son[now][0]=lson; son[now][1]=rson;        Update(now);        return now;    }    inline void Reverse(int l,int r) {splay(l,0),splay(r,l); rev[ls(rs(root))]^=1;}    inline void Init() {root=BuildTree(1,N+2,0);}}int main(){    N=read();    for (int i=2,t=0; i<=N+1; i++) a[i].id=read(),a[i].t=++t;    a[1].id=0; a[N+2].id=N+2;     stable_sort(a+2,a+N+2,cmp);    for (int i=1; i<=N; i++) a[a[i+1].t+1].id=i;//    for (int i=1; i<=N+2; i++) printf("%d ",a[i].id); puts("");    SplayTree::Init();    for (int i=1; i<=N; i++)        {            int loc=SplayTree::GetK(pos[i]);            int x=SplayTree::GetX(i),y=SplayTree::GetX(loc+2);//            printf("%d %d\n",x,y);            SplayTree::Reverse(x,y);            printf("%d%c",loc,i!=N?  :\n);        }    return 0;}

和char哥,龙哥,连坐调splay...感觉智障++

【BZOJ-1552&3506】robotic sort&排序机械臂 Splay