HDU 1890——Robotic Sort（伸展樹）

Robotic Sort

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2511    Accepted Submission(s): 1112

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
        

————————————————————————分割線————————————————————

題目大意：

對你一個序列，每次将第i個位置到第i大的數所在位置 之間的數進行翻轉，輸出第i大的數所在的位置

思路：

對序列離散化為1~n，然後用個數組存儲權值為i，在伸展樹中的節點号為rt，mp[x]=rt;

從小到大将權值為i(第i大)，節點編号為mp[i]旋轉到根，那麼這個數在序列中的位置就是i+size[ch[root][0]]，然後标記rev

注意點：

不用建立邊界點，否則在翻轉操作會把邊界點也一起翻轉，導緻錯誤

寫得蛋碎，自己太弱了....

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#define REP(i,n) for(int i=0;i<(n);++i)
#define Key_value ch[ch[root][1]][0]
#define lson ch[rt][0]
#define rson ch[rt][1]
const int maxn=100010;
using namespace std;
int n;
struct node {
    int x;
    int id;
    bool operator<(const node&cmp) const{
        return ((x<cmp.x)||(x==cmp.x&&id<cmp.id));
    }
} a[maxn];
struct SplayTree{
    int root,tot;
    int ch[maxn][2],pre[maxn],rev[maxn],size[maxn];
    int dis[maxn],mp[maxn],key[maxn];
    void update_rev(int rt) {
        if(!rt) return;
        swap(lson,rson);
        rev[rt]^=1;
    }
    void push_up(int rt) {
        size[rt]=size[lson]+size[rson]+1;
    }
    void push_down(int rt) {
        if(rev[rt]) {
            update_rev(lson);
            update_rev(rson);
            rev[rt]=0;
        }
    }
    void NewNode(int &rt,int f) {
        rt=++tot;
        lson=rson=0;
        pre[rt]=f;
        size[rt]=1;
        rev[rt]=0;
    }
    void build(int &rt,int l,int r,int f) {
        if(l>r) return ;
        int m=(l+r)>>1;
//        cout<<dis[m]<<endl;
        NewNode(rt,f);
        mp[dis[m]]=rt;
        build(lson,l,m-1,rt);
        build(rson,m+1,r,rt);
        push_up(rt);
    }
    void Init() {
        root=tot=0;
        ch[0][0]=ch[0][1]=pre[0]=size[0]=rev[0]=0;
        NewNode(root,0);
        NewNode(ch[root][1],root);
        REP(i,n) {
            scanf("%d",&a[i].x);a[i].id=i;
        }
        sort(a,a+n);
        REP(i,n) dis[a[i].id]=i;
        mp[dis[0]]=1;//注意這兩個操作，沒有建立邊界端點；如果沒有，那麼翻轉的時候會把邊界端點也翻轉，導緻錯誤！！！
        mp[dis[n-1]]=2;//
        build(Key_value,1,n-2,ch[root][1]);
        push_up(ch[root][1]);
        push_up(root);
    }
    void Rotate(int x,int kind) {
        int y=pre[x];
        push_down(y);
        push_down(x);
        ch[y][!kind]=ch[x][kind];
        pre[ch[x][kind]]=y;
        if(pre[y])
            ch[pre[y]][ch[pre[y]][1]==y]=x;
        pre[x]=pre[y];
        ch[x][kind]=y;
        pre[y]=x;
        push_up(y);
    }
    void Splay(int x,int goal) {
        push_down(x);
        while(pre[x]!=goal) {
            if(pre[pre[x]]==goal) {
                Rotate(x,ch[pre[x]][0]==x);
            } else {
                int y=pre[x];
                int kind=ch[pre[y]][0]==y;
                if(ch[y][kind]==x) {
                    Rotate(x,!kind);
                    Rotate(x,kind);
                } else {
                    Rotate(y,kind);
                    Rotate(x,kind);
                }
            }
        }
        push_up(x);
        if(goal==0) root=x;
    }
    void RotateTo(int k,int goal) {
        int rt=root;
        push_down(rt);
        while(size[lson]+1!=k) {
            if(size[lson]>=k) rt=lson;
            else {
                k-=(size[lson]+1);
                rt=rson;
            }
            push_down(rt);
        }
        Splay(rt,goal);
    }
    int Get_kth(int rt,int k) {
        push_down(rt);
        int t=size[lson]+1;
        if(t==k) return rt;
        if(t>k) return Get_kth(lson,k);
        else return Get_kth(rson,k-t);
    }
    void del_root() {
        if(ch[root][1]) {
            int t=root;
            root=ch[root][1];
            //
            RotateTo(1,0);
            ch[root][0]=ch[t][0];
            if(ch[t][0]) pre[ch[t][0]]=root;
        } else root=ch[root][0];

        pre[root]=0;
        push_up(root);
    }
    void solve() {
        REP(i,n) {
            Splay(mp[i],0);
            if(i>0) printf(" ");
            printf("%d",size[ch[root][0]]+i+1);
            update_rev(ch[root][0]);
            del_root();
//            cnt=0;InOrder(root);puts("");
        }
        puts("");
    }
}spt;
int main()
{
    while(scanf("%d",&n),n) {
        if(n==1) {scanf("%*d");printf("1\n");continue;}
        spt.Init();
        spt.solve();
    }
    return 0;
}