文章标题

poj 2299 Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536K

Total Submissions: 53779 Accepted: 19767

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

解题思路：利用分治的方法来排序，这样是所有排序的算法里边的最省时的一个，已经是世界都公认的，排完序的过程就可以计算最后的counts的值；

我的代码就在下边，不过很抱歉，这个头文件不知道在这个markdown编译器里边怎么排版，所以头文件有点怪怪的，抱歉啊，具体的题解在代码里边有，读者可以看看，笔者就不多说了。

Waterloo local 2005.02.05

#include<iostream>#include<cstdio>#include<cstring>

using namespace std;

__int64 counts=0;//代表counts是long long 型数据

int a[500005],A[500005],n;

void Sort(int from,int mid,int to)

{

int i=from,j=mid+1,k=from;

while(i<=mid&&j<=to)

if(a[i]<=a[j]){A[k++]=a[i++];}

else {A[k++]=a[j++];counts+=j-k;}

while(i<=mid)A[k++]=a[i++];//后边三步其实在这道题都不用写，因为并没有让给a数组排序

while(j<=to)A[k++]=a[j++];

for(int i=from;i<=to;i++)

a[i]=A[i];

}

void Merge_Sort(int from,int to)

{

if(from>=to)return;

int mid=(from+to)>>1;//用位移符号比除号速度更快，更省时

Merge_Sort(from,mid);//分治左边

Merge_Sort(mid+1,to);//分治右边

Sort(from,mid,to);//一定要记得合起来；

}

int main()

{

while(~scanf("%d",&n)&&n)

{

for(int i=0;i<n;i++)

scanf("%d",&a[i]);

counts=0;

Merge_Sort(0,n-1);

printf("%I64d\n",counts);//只有64位才不会wa！！！！！

}

}