Poj 2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 41876		Accepted: 15208

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

#include<iostream>
#include<stdio.h>
using namespace std;
int a[500002];

//递归2路归并排序
/*void Merge(long long a[],long long b[],int s,int m,int t)
{
int i=s,j=m+1,k=s;
while((i<=m)&&(j<=t))
{
if(a[i]<=a[j]) b[k++]=a[i++];
else b[k++]=a[j++];
}
while(i<=m) b[k++]=a[i++];
while(j<=t) b[k++]=a[j++];
}
void MSort(long long a[],long long b[],int s,int t ,int size)
{
int m;
long long c[size];
if(s==t) b[s]=a[t];
else
{
m=(s+t)/2;
MSort(a,c,s,m,size);
MSort(a,c,m+1,t,size);
Merge(c,b,s,m,t);
}
}
void MergeSort(long long a[],int size)
{
MSort(a,a,0,size-1,size);
}*/

// 非递归合并排序
/*template<class T>
void Merge(T a[],T b[],int s,int m,int t)
{
int i=s,j=m+1,k=s;
while(i<=m&&j<=t)
{
if(a[i]<a[j]) b[k++]=a[i++];
else b[k++]=a[j++];
}
while(i<=m) b[k++]=a[i++];
while(j<=t) b[k++]=a[j++];
}
template<class T>
void MergePass(T a[],T b[],int s,int t)
{
int i;
for(i=0;i+2*s<=t;i=i+2*s)
{
Merge(a,b,i,i+s-1,i+2*s-1);
}
//剩下的元素个数少于2s
if(i+s<t) Merge(a,b,i,i+s-1,t);
else for(int j=i;j<=t;j++) b[j]=a[j];
}
template<class T>
void MergeSort(T a[],int n)
{
T *b=new T
;
//T b
;
int s=1;
while(s<n)
{
MergePass(a,b,s,n);
s+=s;
MergePass(b,a,s,n);
s+=s;
}
}*/

long long count=0;
//求逆序对
void Merge(int a[],int b[],int s,int m,int t)
{
int i=s,j=m+1,k=s;
//int count=0;
while(i<=m&&j<=t)
{
if(a[i]<=a[j]) b[k++]=a[i++];
else
{
b[k++]=a[j++];
count+=m-i+1;
}

}
while(i<=m) b[k++]=a[i++];
while(j<=t) b[k++]=a[j++];
//return count;
}
void MergePass(int a[],int b[],int s,int t)
{
int i;
//int count=0;
for(i=0;i+2*s<=t;i=i+2*s)
{
Merge(a,b,i,i+s-1,i+2*s-1);
}
//剩下的元素个数少于2s
if(i+s<t) Merge(a,b,i,i+s-1,t-1);
else for(int j=i;j<=t;j++) b[j]=a[j];
//return count;
}
void MergeSort(int a[],int n)
{
int *b=new int
;
//int b
;
//int count=0;
int s=1;
while(s<n)
{
MergePass(a,b,s,n);
s+=s;
MergePass(b,a,s,n);
s+=s;
}
}

int main()
{

//测试排序正确性
/*int a[10];
for(i=0;i<10;i++) a[i]=10-i;
for(i=0;i<10;i++) cout<<a[i]<<" ";
cout<<endl;
cout<<MergeSort(a,10)<<endl;
for(i=0;i<10;i++) cout<<a[i]<<" ";*/

int size;
int i;
while((cin>>size)&&size!=0)
{
count=0;
for(i=0;i<size;i++)
{
scanf("%lld",&a[i]);
}
MergeSort(a,size);
printf("%lld\n",count);
//测试排序正确性
//for(i=0;i<size;i++) printf("%lld ",a[i]);
}
//system("pause");
return 1;
}