http://acm.hdu.edu.cn/showproblem.php?pid=1890
Robotic Sort
Problem Description
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.
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.
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.
0
#include"stdio.h"
#include"iostream"
#include"queue"
#include"string.h"
#include"map"
#include"stdlib.h"
#include"algorithm"
#include"string"
#define M 1000005
#define inf 0x3f3f3f3f
using namespace std;
int a[M];
struct P
{
int val,pri,key;
}p[M];
int top,root;
int cmp(P a,P b)
{
if(a.val==b.val)
return a.pri<b.pri;
return a.val<b.val;
}
int son[M][],fa[M],num[M],flip[M],child[M];
struct Text
{
queue<int>q;
void init(int n)
{
top=n+;
for(int i=;i<=n+;i++)
{
num[i]=;
son[i][]=son[i][]=-;
fa[i]=-;
flip[i]=;
}
}
void Rotate(int x,int k)
{
int y=fa[x]; son[y][^k]=son[x][k];
if(son[y][^k]!=-)
fa[son[y][^k]]=y;
push_up(y);
if(son[fa[y]][]==y)
son[fa[y]][]=x;
else
son[fa[y]][]=x;
fa[x]=fa[y]; son[x][k]=y;
fa[y]=x;
}
void splay(int x,int f)
{
if(x==-)return;
while(fa[x]!=f)
{
int y=fa[x];
int z=fa[y];
if(z==f)
{
if(son[y][]==x)
Rotate(x,);
else
Rotate(x,);
}
else
{
if(son[z][]==y)
{
if(son[y][]==x)
{
Rotate(y,);
Rotate(x,);
}
else
{
Rotate(x,);
Rotate(x,);
}
}
else
{
if(son[y][]==x)
{
Rotate(y,);
Rotate(x,);
}
else
{
Rotate(x,);
Rotate(x,);
}
}
}
}
if(f==top)
root=x;
}
void RotateTo(int k,int f)
{
k++;
int x=root;
while()
{
push_down(x);
int temp=getNum(son[x][])+;
if(k==temp)break;
else if(k<temp)
x=son[x][];
else
{
k-=temp;
x=son[x][];
}
}
splay(x,f);
push_up(x);
}
int getOrder(int x)
{
int y=x;
while(fa[y]!=top)
{
int f=fa[y];
child[f]=y;
y=fa[y];
}
y=root;
while(y!=x)
{
int z=child[y];
push_down(z);
y=child[y];
}
int k=getNum(son[x][])+;
while(fa[x]!=top)
{
int y=fa[x];
int temp=getNum(son[y][])+;
if(son[y][]==x)
x=fa[x];
else
{
k+=temp;
x=fa[x];
}
}
return k-;
}
void Reversal(int x)
{
if(x==-)return;
int y=son[x][];
son[x][]=son[x][];
son[x][]=y;
}
int getNum(int x)
{
if(x==-)return ;
return num[x];
}
void push_up(int x)
{
num[x]=getNum(son[x][])+getNum(son[x][])+;
}
void push_down(int x)
{
if(x==-)return;
if(flip[x])
{
flip[x]^=;
Reversal(x);
if(son[x][]!=-)
flip[son[x][]]^=;
if(son[x][]!=-)
flip[son[x][]]^=;
}
}
void creat(int l,int r,int k,int f)
{
if(l>r)return;
int mid=(l+r)/;
if(f==top)
root=mid;
son[f][k]=mid;
fa[mid]=f;
creat(l,mid-,,mid);
creat(mid+,r,,mid);
push_up(mid);
}
void dfs(int x,int n)
{
if(x==-)return;
push_down(x);
dfs(son[x][],n);
if(x>=&&x<=n)
q.push(a[x]);
dfs(son[x][],n);
}
void output(int n)
{
dfs(root,n);
printf("%d",q.front());
q.pop();
while(!q.empty())
{
printf(" %d",q.front());
q.pop();
}
puts("");
}
}; int main()
{
int n;
while(scanf("%d",&n),n)
{
Text text;
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
map<int,int>mp;
p[].val=-inf;
p[].pri=;
p[].key=;
for(int i=;i<=n;i++)
{
p[i].val=a[i];
p[i].pri=mp[a[i]];
p[i].key=i;
mp[a[i]]++;
}
p[n+].val=inf;
p[n+].pri=;
p[n+].key=n+;
sort(p,p+n+,cmp);
text.init(n);
text.creat(,n+,,top);
for(int i=;i<=n;i++)
{
int l=i-;
int key=p[i].key;
int r=text.getOrder(key)+;
text.RotateTo(l,top);
text.RotateTo(r,root);
flip[son[son[root][]][]]^=;
//text.output(n);
if(i==)
printf("%d",r-);
else
printf(" %d",r-);
}
puts("");
}
return ;
}