poj 2299 Ultra-QuickSort(树状数组）

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 ：多组输入，输入n，n==0结束，接着输入n个数。把输入的数排好序需要几步。
思路：由于输入的数不是从1开始公差为1的递增序列，先用两个快排转换一下，代入树状数组就ok了！
ac代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
long long d1[5000000];
int n;
struct node                                      //g存数，h存下标
{
    int g,h;
} s[5000000];
int cmp(struct node a,struct node b)             //控制排序的方向从大到小
{
    return a.g>b.g;
}
int cmp1(struct node a,struct node b)            //控制排序方向从小到大
{
    return a.h<b.h;
}
int lowbit(int x)                               
{
    return x&(-x);
}
long long sum(int x)                             //树状数组求和，这题是求x前面有多少个数，因为把所有数转换成公差为1的递增序列，求出的和就是答案
{
    long long res=0;
    while(x>0)
    {
        res+=d1[x];
        x-=lowbit(x);
    }
    return res;
}
void add(int x)                                  //更新树状数组
{
    while(x<=n)
    {
        d1[x]++;
        x+=lowbit(x);
    }
}
int main()
{
    int a,b,i,j;
    long long num;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        break;
        num=0;
        memset(d1,0,sizeof(d1));
        for(i=1; i<=n; i++)
        {
            scanf("%d",&s[i].g);
            s[i].h=i;
        }
        sort(s+1,s+n+1,cmp);                          //按g值的大小排序
        for(i=1; i<=n; i++)
            s[i].g=i;
        sort(s+1,s+n+1,cmp1);                         //按h排序 ，就是原来的下标
        for(i=1;i<=n;i++)
        {
            num+=sum(s[i].g);                         //把这个数前面有的数的个数相加，就是答案
            add(s[i].g);
        }
        printf("%lld
",num);
    }
}