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

59105431230

Sample Output

60

解题思路：

题目大意是给一个数列，相邻两个进行交换，使之按从小到大排序，问最少交换几次。该题和之前的做的Janan是一个类型的，都是求逆序对。唯一难点就是数据特别大，数列中的元素值可以达到999999999，树状数组不可能开这么大。但由于数列最多有500000个数，所以可以进行离散化处理，把数列中的元素压缩到1-500000之间。离散化就是将输入的值与下标相对应，可以用结构体实现，然后对输入的值进行从小到大排序，再用一个数组去存储其下标的值。答案会超int范围，得用Int64存储。

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn = 500005;__int64 c[maxn];struct node{    int a, b;  // a存储输入的值，b存储其坐标}p[maxn];bool cmp(node v, node s){    return v.a < s.a;}int lowbit(int a){    return a & (-a);}void Update(int a){    while(a < maxn)    {        c[a] += 1;        a += lowbit(a);    }}__int64 Sum(int a){    __int64 sum = 0;    while(a > 0)    {        sum += c[a];        a -= lowbit(a);    }    return sum;}int main(){    int n, a[maxn];    __int64 ans;    while(scanf("%d", &n) && n)    {        ans = 0;        memset(c, 0, sizeof(c));        for(int i = 1; i <= n; i++)        {            scanf("%d", &p[i].a);            p[i].b = i;        }        sort(p + 1, p + n + 1, cmp);        for(int i = 1; i <= n; i++)  // 离散化处理            a[p[i].b] = i;        for(int i = 1; i <= n; i++)        {            ans += i - Sum(a[i]) - 1;  // 要-1，因为算的是输入该值之前的元素个数            Update(a[i]);        }        printf("%I64d\n", ans);    }    return 0;}