POJ 2299 Ultra-QuickSort 树状数组

                                                                                              Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 32009		Accepted: 11395

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 sequence9 1 0 5 4 ,Ultra-QuickSort produces the output0 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

代码 ：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std ;
#define M 500010
int xt[M*2] , n ;
int a[M] ;
struct node{ int w , id ; } qe[M] ;// 用结构体储存 原始数据 id 是数据原始的位置
int cmp( node a , node b ){ 
	return a.w < b.w ;
}
int lowbit( int x ){
	return x & ( -x ) ;
}
void update( int x ){
	while( x <= n ){
		xt[x] += 1 ;
		x += lowbit(x) ;
	}

}
int sum( int x ){
	int s = 0 ;
	while( x > 0 ){
		s += xt[x] ;
		x -= lowbit(x) ;
	}
	return s ;
}
int main()
{
	int i ;
	long long mun ;
	while( cin >> n ){
		if( n == 0 ) break ;
		mun = 0 ;
		memset( xt , 0 , sizeof(xt) ) ;
		for( i = 1; i <= n ;i++){
			scanf( "%d" , &qe[i].w ) ;
			qe[i].id = i ;
		} 
		sort( qe + 1 , qe + 1 + n , cmp ) ;
		for( i = 1 ; i <= n ; i++){
             a[qe[i].id] = i ; // 对数据进行离散化 例如 原来是 9 0 1 2 处理后是 4 1 2 3 
		}
		for( i = 1 ; i <= n ; i++){
			update(a[i] ) ;
			mun += ( i - sum(a[i] - 1 ) - 1 ) ;
		}
		cout << mun << endl ;
	}
}