hdu1890 Robotic Sort (splay+区间翻转单点更新)

Problem Description

Somewhere deep in the Czech Technical University buildings, there are laboratories for examining mechanical and electrical properties of various materials. In one of yesterday’s presentations, you have seen how was one of the laboratories changed into a new multimedia lab. But there are still others, serving to their original purposes. 

In this task, you are to write software for a robot that handles samples in such a laboratory. Imagine there are material samples lined up on a running belt. The samples have different heights, which may cause troubles to the next processing unit. To eliminate such troubles, we need to sort the samples by their height into the ascending order. 

Reordering is done by a mechanical robot arm, which is able to pick up any number of consecutive samples and turn them round, such that their mutual order is reversed. In other words, one robot operation can reverse the order of samples on positions between A and B. 

A possible way to sort the samples is to find the position of the smallest one (P1) and reverse the order between positions 1 and P1, which causes the smallest sample to become first. Then we find the second one on position P and reverse the order between 2 and P2. Then the third sample is located etc. 



The picture shows a simple example of 6 samples. The smallest one is on the 4th position, therefore, the robot arm reverses the first 4 samples. The second smallest sample is the last one, so the next robot operation will reverse the order of five samples on positions 2–6. The third step will be to reverse the samples 3–4, etc. 

Your task is to find the correct sequence of reversal operations that will sort the samples using the above algorithm. If there are more samples with the same height, their mutual order must be preserved: the one that was given first in the initial order must be placed before the others in the final order too.

 

Input

The input consists of several scenarios. Each scenario is described by two lines. The first line contains one integer number N , the number of samples, 1 ≤ N ≤ 100 000. The second line lists exactly N space-separated positive integers, they specify the heights of individual samples and their initial order. 

The last scenario is followed by a line containing zero.

 

Output

For each scenario, output one line with exactly N integers P1 , P1 , . . . PN ,separated by a space.
Each Pi must be an integer (1 ≤ Pi ≤ N ) giving the position of the i-th sample just before the i-th reversal operation. 

Note that if a sample is already on its correct position Pi , you should output the number Pi anyway, indicating that the “interval between Pi and Pi ” (a single sample) should be reversed. 

 

Sample Input

6 3 4 5 1 6 2 4 3 3 2 1 0

 

Sample Output

4 6 4 5 6 6

4 2 4 4

题意：给一个长度为n的数列，每次选取值最小的元素并翻转前面的数列，然后删除这个元素。请在每次操作之前输出这个最小元素的位置。

思路：先对原来的序列排序，然后预处理出第i大的数在树上的节点编号，然后每一次把第i大的节点旋到根节点，那么答案就是i+sz[ch[rt][0] ],然后删除这个节点。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define lson th<<1
#define rson th<<1|1
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 100050
#define Key_value ch[ch[root][1]][0]
int n;
struct edge{
    int idx,num;
}a[maxn];
int mp[maxn],mp1[maxn];
bool cmp(edge a,edge b){
    if(a.num==b.num)return a.idx<b.idx;
    return a.num<b.num;
}

int cnt,rt;
int pre[maxn],ch[maxn][2],sz[maxn],rev[maxn];

void newnode(int &x,int father)
{
    x=++cnt;
    pre[x]=father;ch[x][0]=ch[x][1]=0;sz[x]=1;rev[x]=0;
}
void update_rev(int x)
{
    if(x==0)return;       //!!!
    rev[x]^=1;
    swap(ch[x][0],ch[x][1]);
}
void pushdown(int x)
{
    int y;
    if(rev[x]){
        update_rev(ch[x][0]);
        update_rev(ch[x][1]);
        rev[x]=0;
    }
}
void pushup(int x)
{
    sz[x]=sz[ch[x][0] ]+sz[ch[x][1] ]+1;
}

void build(int &x,int l,int r,int father)
{
    if(l>r)return;
    int mid=(l+r)/2;
    newnode(x,father);mp1[mp[mid] ]=cnt;
    build(ch[x][0],l,mid-1,x);
    build(ch[x][1],mid+1,r,x);
    pushup(x);
}

void init()
{
    cnt=rt=0;
    pre[rt]=ch[rt][0]=ch[rt][1]=sz[rt]=rev[rt]=0;
    build(rt,1,n,0);
}


void rotate(int x,int p)
{
    int y=pre[x];
    pushdown(y);pushdown(x);
    ch[y][!p]=ch[x][p];
    pre[ch[x][p] ]=y;
    if(pre[y])ch[pre[y] ][ch[pre[y] ][1]==y ]=x;
    pre[x]=pre[y];
    ch[x][p]=y;
    pre[y]=x;
    pushup(y);pushup(x);
}
void splay(int x,int goal)
{
    pushdown(x);
    while(pre[x]!=goal){
        if(pre[pre[x] ]==goal){
            pushdown(pre[x]);pushdown(x);
            rotate(x,ch[pre[x]][0]==x);
        }
        else{
            int y=pre[x];int z=pre[y];
            pushdown(z);pushdown(y);pushdown(x);
            int p=ch[pre[y] ][0]==y;
            if(ch[y][p]==x )rotate(x,!p);
            else rotate(y,p);
            rotate(x,p);
        }
    }
    if(goal==0)rt=x;
    pushup(x);
}
void del()
{
    if(ch[rt][0]==0 ){
        rt=ch[rt][1];
        pre[rt]=0;
    }
    else{
        int y=ch[rt][0];
        int x=ch[rt][1];
        pushdown(y);
        while(ch[y][1]){
            y=ch[y][1];pushdown(y);
        }
        splay(y,rt);
        ch[y][1]=x;
        pre[x]=y;
        rt=y;
        pre[rt]=0;
        pushup(rt);
    }
}

int main()
{
    int m,i,j;
    while(scanf("%d",&n)!=EOF && n!=0)
    {
        for(i=1;i<=n;i++){
            scanf("%d",&a[i].num);
            a[i].idx=i;
        }
        sort(a+1,a+1+n,cmp);
        for(i=1;i<=n;i++)mp[a[i].idx ]=i;

        init();
        for(i=1;i<n;i++){
            splay(mp1[i],0);
            update_rev(ch[rt][0]);
            printf("%d ",i+sz[ch[rt][0]]);
            del();
        }
        printf("%d
",n);
    }
    return 0;
}