poj2299 查找逆序数 归并

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 49032		Accepted: 17936

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
查找逆序数

/*
由于本题实际上就是要求逆序对（即满足ia[j]的数对）的个数。而我们再回顾一下归并排序的过程：

假设回溯到某一步，后面的两部分已经排好序（就是说当前需要归并的两个部分都是分别有序的），假设这两个序列为

序列a1：2 3 5 9  

序列a2：1 4 6 8

此时我们的目的就是要将a1和a2合并为一个序列。

由于在没排序前a2序列一定全部都是在a1序列之后的，当我们比较a2的1与a1的2时，发现1<2按照归并的思想就会先记录下a2的1，而这里实际上就是对冒泡排序的优化，冒泡是将a2的1依次与a1的9,5,3,2交换就需要4次，而归并却只有一次就完成了，要怎么去记录这个4呢，实际上由于1比2小而2后面还有4个数，也就是说那我的结果就必须要+4，也就是记录a1序列找到第一个比a2某一个大的数，他后面还余下的数的个数就是要交换的次数。

同时我们看a2的4时，a1中第一个比它大的数是5,5之后共有两个数，那结果就+2,。依次下去就可以计算出结果。但是由于我们任然没有改变归并排序的过程。所以复杂度还是O(nlogn)

此题有一坑就是结果会超int32,，用__int64*/

/*

#include
#include
#include
#include

using namespace std;

int a[500050],Arr[500050];
long long int Sort(int low,int mid,int high)
{
	int i=low,j=mid+1,k=low;
	long long ans=0;
	while(i<=mid&&j<=high)
	{
		if(a[i]<=a[j])
			Arr[k++]=a[i++];
		else
		{
			Arr[k++]=a[j++];
			ans+=j-k;
		}

	}
	while(i<=mid)
	{
		Arr[k++]=a[i++];
	}
	while(j<=high)
	{
		Arr[k++]=a[j++];
	}
	for(int i=low;i<=high;i++)
	{
		a[i]=Arr[i];
	}
	return ans;
}
long long  int MergeSort(int low,int high)
{
	if(low>=high)     //勿漏
		return 0;
	long long num=0;
	int mid=(low+high)/2;
	num+=MergeSort(low,mid);
	num+=MergeSort(mid+1,high);
	num+=Sort(low,mid,high);
	return num;
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		if(n==0)
			break;
		for(int i=0;i

#include
#include
#include
#include

using namespace std;

int a[500050],Arr[500050];
long long num=0;
long long int Sort(int low,int mid,int high)
{
	int i=low,j=mid+1;
	int k=low;

	while(i<=mid&&j<=high)
	{
		if(a[i]<=a[j])      //
			Arr[k++]=a[i++];
		else
		{
			Arr[k++]=a[j++];
			num+=j-k;      //逆序数的统计
		}

	}
	while(i<=mid)
	{
		Arr[k++]=a[i++];
	}
	while(j<=high)
	{
		Arr[k++]=a[j++];
	}
	for(int i=low;i<=high;i++)
	{
		a[i]=Arr[i];
	}
}
long long  int MergeSort(int low,int high)
{
	if(low>=high)     //勿漏
		return 0;

	int mid=(low+high)/2;
	 MergeSort(low,mid);
	 MergeSort(mid+1,high);
	 Sort(low,mid,high);

}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		if(n==0)
			break;
		for(int i=0;i