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

59 1 0 5 431 2 30

Sample Output

60

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\n",ans);}return 0;}