poj2299：Ultra-QuickSort（树状数组+离散化）

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

题意：给你一个n个整数组成的序列，每次只能交换相邻的两个元素，问你最少要进行多少次交换才能使得整个整数序列上升有序。  普通方法肯定超时  用树状数组+离散化

因为数据大小范围为0~999,999,999，若用此数作为数组下标肯定会超出内存限制，所以采用离散化的方法先将数据范围缩小

输入的数据：9 1 0 5 4  
  
排序的数据：0 1 4 5 9  
排序的编码：3 2 5 4 1  
离散化之后：1 2 3 4 5  
  
最终的编码：1 2 3 4 5  
最终的数据：5 2 1 4 3   
  
树状数组：  
输入一个数据看看前面比他大的数据有几个；求他们的和  

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int n; 
struct node
{
	int val,zb;
}a[500001];//初始数组 
int b[500001];//离散化后新数组
int id[500001];//离散化后的新坐标 
bool cmp(node a,node b)
{
	return a.val <b.val ;
}
int lowbit(int x)
{
	return x&-x;
}
void add(int x,int d)
{
	while(x<=n)
	{
			b[x]+=d;
     	x+=lowbit(x);
	}
}
int sum(int x)
{
	int s=0;
	while(x>0)
	{
		s+=b[x];
		x-=lowbit(x);
	}
	return s;
 } 
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
		{
			break;
		}
	  for(int i=1;i<=n;i++)
	  {
	  	b[i]=0;
	  	scanf("%d",&a[i].val );
	  	a[i].zb =i;
	  }
	  sort(a+1,a+n+1,cmp);
	  for(int i=1;i<=n;i++)
	  {
	  	id[a[i].zb ]=i;
	  }
	  long long ans=0;
	  for(int i=1;i<=n;i++)
	  {
	  	add(id[i],1);
	  	ans+=(i-sum(id[i]));
	  }
	  printf("%lld
",ans);
	}
	return 0;
}