HDU 1890 Robotie Sort

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

-------------------------------------------------------------------------------------------分割线------------------------------------------------------------------------------------------------------

题目大意：

　　给出一段序列，长度为N。要将这段序列进行排序。排序的方式是通过每次找出第i大数与它本来的位置一段序列进行反转，相当于一次排序、也就相当于反转N次，要求每次输出反转前第i大位置的数的位置。

　　每组输入数据包括两行，第一行输入序列长度N，第二行输入该序列。

　　输出数据包括N个数字，Pi~PN。Pi表示第i次反转前第i大的位置。

 

AC代码如下：

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<string.h>
  4 using namespace std;
  5 #define N 500006
  6 #define lc (tr[id].c[0])
  7 #define rc (tr[id].c[1])
  8 #define key (tr[tr[root].c[1]].c[0])
  9 struct tree
 10 {
 11     int fa,sum,c[2],lz,v;
 12 }tr[N];
 13 struct point
 14 {
 15     int v,id;
 16     bool operator<(const point a)const
 17     {
 18         if(a.v==v)return id<a.id;
 19         else return v<a.v;
 20     }
 21 }so[N/5];
 22 int tot,root,n;
 23 int xia[N];
 24 int newpoint(int d,int f)
 25 {
 26     tr[tot].sum=1;
 27     tr[tot].v=d;
 28     tr[tot].c[0]=tr[tot].c[1]=-1;
 29     tr[tot].lz=0;
 30     tr[tot].fa=f;
 31     return tot++;
 32 }
 33 void push(int id)
 34 {
 35     int lsum,rsum;
 36     if(lc==-1)lsum=0;
 37     else lsum=tr[lc].sum;
 38     if(rc==-1)rsum=0;
 39     else rsum=tr[rc].sum;
 40     tr[id].sum=lsum+rsum+1;
 41 }
 42 int build(int l,int r,int v)
 43 {
 44     if(r<l)return -1;
 45     int mid=(r+l)>>1;
 46     int ro=newpoint(mid,v);
 47     xia[mid]=ro;
 48     tr[ro].c[0]=build(l,mid-1,ro);
 49     tr[ro].c[1]=build(mid+1,r,ro);
 50     push(ro);
 51     return ro;
 52 }
 53 void lazy(int id)
 54 {
 55     if(tr[id].lz)
 56     {
 57         swap(lc,rc);
 58         tr[lc].lz^=1;
 59         tr[rc].lz^=1;
 60         tr[id].lz=0;
 61     }
 62 }
 63 
 64 void xuanzhuan(int x,int k)
 65 {
 66     if(tr[x].fa==-1)return ;
 67     int fa=tr[x].fa;
 68     int w;
 69     lazy(fa);
 70     lazy(x);
 71     tr[fa].c[!k]=tr[x].c[k];
 72     if(tr[x].c[k]!=-1)tr[tr[x].c[k]].fa=fa;
 73     tr[x].fa=tr[fa].fa;
 74     tr[x].c[k]=fa;
 75     if(tr[fa].fa!=-1)
 76     {
 77         w=tr[tr[fa].fa].c[1]==fa;
 78         tr[tr[fa].fa].c[w]=x;
 79     }
 80     tr[fa].fa=x;
 81     push(fa);
 82     push(x);
 83 }
 84 
 85 void splay(int x,int goal)
 86 {
 87     if(x==-1)return ;
 88     lazy(x);
 89     while(tr[x].fa!=goal)
 90     {
 91         int y=tr[x].fa;
 92         lazy(tr[y].fa);
 93         lazy(y);
 94         lazy(x);
 95         bool w=(x==tr[y].c[1]);
 96         if(tr[y].fa!=goal&&w==(y==tr[tr[y].fa].c[1]))xuanzhuan(y,!w);
 97         xuanzhuan(x,!w);
 98     }
 99     if(goal==-1)root=x;
100     push(x);
101 }
102 int next(int id)
103 {
104     lazy(id);
105     int p=tr[id].c[1];
106     if(p==-1)return id;
107     lazy(p);
108     while(tr[p].c[0]!=-1)
109     {
110         p=tr[p].c[0];
111         lazy(p);
112     }
113     return p;
114 }
115 int main()
116 {
117     while(scanf("%d",&n),n)
118     {
119         for(int i=1;i<=n;i++)
120         {
121             scanf("%d",&so[i].v);
122             so[i].id=i;
123         }
124         sort(so+1,so+n+1);
125         so[0].id=0;
126         tot=0;
127         int d,l;
128         root=build(0,n+1,-1);
129         for(int i=1;i<=n;i++)
130         {
131             int ro=xia[so[i].id];
132             int ne;
133             splay(ro,-1);
134             d=tr[tr[root].c[0]].sum;
135             l=xia[so[i-1].id];
136             ne=next(ro);
137             splay(l,-1);
138             splay(ne,root);
139             lazy(root);
140             lazy(tr[root].c[1]);
141             tr[key].lz^=1;
142             if(i!=1)printf(" ");
143             printf("%d",d);
144         }
145         printf("
");
146     }
147     return 0;
148 }

 题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=1890

（本人蒟蒻，代码也是看书上敲的，以后应该会慢慢成长吧）