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Ultra-QuickSort（树状数组 + 离散化）

2024-07-14 20:39:47   219人阅读

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

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;
}




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