POJ 2299 Ultra-QuickSort 树状数组
Ultra-QuickSort
Time Limit: 7000MS | Memory Limit: 65536K | |
Total Submissions: 32009 | Accepted: 11395 |
Description
![](https://www.cnblogs.com/20120125llcai/admin/images/2299_1.jpg)
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 ; } }