POJ 2299 Ultra-QuickSort

题目链接 : POJ2299
[align=center]Ultra-QuickSort[/align]

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 32350		Accepted: 11526

Description


In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements
until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,

Ultra-QuickSort produces the output 
0 1 4 5 9 .

Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.
Input
The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999,
the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.
Output
For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.
Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

Source
Waterloo local 2005.02.05
 
唉。。

 

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<vector>
#include<iostream>
#include<map>
#include<ctime>
using namespace std;
#define INF 0x7fffffff

int num[500005],temp[500005];
long long ans;
void pai(int sta,int end)
{
int s=sta,e=end-1,con=0;
for(int i=end-1;i>sta;i--)
{
if(num[i]>num[sta])
{
temp[e--]=num[i];
ans+=con;
}
else
{
temp[s++]=num[i];
con++;
}
}
ans+=con;
temp[e]=num[sta];
for(int i=sta;i<e;i++)
num[i]=temp[e-i+sta-1];
for(int i=e;i<end;i++)
num[i]=temp[i];
if(end-e>2) pai(e+1,end);
if(e-sta>1) pai(sta,e);
}
int main()
{
int n;
while(scanf("%d",&n))
{
if(n==0) break;
ans=0;
for(int i=0;i<n;i++)
scanf("%d",num+i);
pai(0,n);
printf("%lld\n",ans);
}
return 0;
}