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("\n");
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）