HDU 1890 Splay区间翻转

标签：splay   bst   

D - Robotic Sort

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1890

Appoint description:  System Crawler  (2014-11-27)

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 

 

用翻转对区间排序，建一个splay为1,2，...n，搞一个结构体数组A[maxn]，first保存key，second保存在初始数组中的下标，按照key排个序，则A[i].second就是第i大的元素在splay中的编号，每次将其旋转到跟，比他小的元素加上i就是此时他在元素中的位置，输出后删除树根(因为这一部分已经有序啦,不删除会对后面的操作有影响)。翻转最好一次翻转两层，开始我只翻转一层就超时了。

代码：

/*************************************************************************
    > File Name: Spaly.cpp
    > Author: acvcla
    > QQ:
    > Mail: acvcla@gmail.com
    > Created Time: 2014Äê11ÔÂ16ÈÕ ÐÇÆÚÈÕ 00ʱ14·Ö26Ãë
 ************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 100;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
int ch[maxn][2],pre[maxn],siz[maxn],rev[maxn];
int root,tot;
struct  node
{
	int first,second;
	bool operator < (const node &a)const{
		if(first==a.first)return second<a.second;
		return first<a.first;
	}
};
node A[maxn];
inline void newnode(int &x,int fa,int id)
{
	x=id;
	siz[x]=1;
	pre[x]=fa;
	ch[x][1]=ch[x][0]=0;
}
void update_rev(int r)  
{  
    if(!r)return;  
    swap(ch[r][0],ch[r][1]);  
    rev[r]^=1;  
}
void push_down(int r)  
{  
    if(rev[r])  
    {  
        update_rev(ch[r][0]);  
        update_rev(ch[r][1]);  
        rev[r]=0;  
    }  
}
inline void push_up(int x)
{
	siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
}
inline 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;
	ch[x][kind]=y;

	if(pre[y])ch[pre[y]][ch[pre[y]][1]==y]=x;
	pre[x]=pre[y];
	pre[y]=x;
	push_up(y);
	push_up(x);
}
inline void Splay(int x,int goal)
{
	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);
			}
		}
	}
	if(goal==0)root=x;
}
int Get_max(int x)
{
	push_down(x);
	while(ch[x][1]){
		x=ch[x][1];
		push_down(x);
	}
	return x;
}
inline void remove_root(){
	push_down(root);
	if(!ch[root][0]){
		root=ch[root][1];
	}else{
		int t=Get_max(ch[root][0]);
		Splay(t,root);
		ch[t][1]=ch[root][1];
		pre[ch[root][1]]=t;
		root=t;
	}
	pre[root]=0;
	push_up(root);
}
inline void built(int &x,int L,int R,int fa)
{
	if(L>R)return;
	int mid=(R+L)>>1;
	newnode(x,fa,mid);
	built(ch[x][0],L,mid-1,x);
	built(ch[x][1],mid+1,R,x);
	push_up(x);
}
inline void init(int n)
{
	pre[0]=siz[0]=0;
	root=tot=ch[0][0]=ch[0][1]=0;
	for(int i=1;i<=n;i++){
		scanf("%d",&A[i].first);
		A[i].second=i;
	}
	sort(A+1,A+1+n);
	built(root,1,n,0);
}
int main(int argc, char const *argv[])
{
	int n;
	while(~scanf("%d",&n)&&n){
		init(n);
		rep(i,1,n-1){
			Splay(A[i].second,0);
			update_rev(ch[root][0]);
			printf("%d ",i+siz[ch[root][0]]);
			remove_root();
		}
		printf("%d\n",n);
	}
	return 0;
}

HDU 1890 Splay区间翻转

标签：splay   bst   

原文地址：http://blog.csdn.net/acvcla/article/details/41660861