Ultra-QuickSort（樹狀數組 + 離散化）

標籤：樹狀數組

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

解題思路：

題目大意是給一個數列，相鄰兩個進行交換，使之按從小到大排序，問最少交換幾次。該題和之前的做的Janan是一個類型的，都是求逆序對。唯一痛點就是資料特別大，數列中的元素值可以達到999999999，樹狀數組不可能開這麼大。但由於數列最多有500000個數，所以可以進行離散化處理，把數列中的元素壓縮到1-500000之間。離散化就是將輸入的值與下標相對應，可以用結構體實現，然後對輸入的值進行從小到大排序，再用一個數組去儲存其下標的值。答案會超int範圍，得用Int64儲存。

AC代碼：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn = 500005;__int64 c[maxn];struct node{    int a, b;  // a儲存輸入的值，b儲存其座標}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++)  // 離散化處理            a[p[i].b] = i;        for(int i = 1; i <= n; i++)        {            ans += i - Sum(a[i]) - 1;  // 要-1，因為算的是輸入該值之前的元素個數            Update(a[i]);        }        printf("%I64d\n", ans);    }    return 0;}