R - POJ 2299 树状数组求逆序对

最新推荐文章于 2021-12-05 21:54:16 发布

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本文探讨了一种特定的排序算法Ultra-QuickSort，该算法通过交换相邻元素来对序列进行排序。文章详细介绍了算法的工作原理，以及如何计算排序过程中所需的最小交换次数。通过分析逆序数的概念，提供了一个有效的解决方案。

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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

Sample Output

6
0

题意：给出长度为n的序列，每次只能交换相邻的两个元素，问至少要交换几次才使得该序列为递增序列。

思路：一个乱序序列的逆序数 = 在只允许相邻两个元素交换的条件下, 得到有序序列的交换次数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 500010;
int num[maxn], tree[maxn];
int n;
long long ans = 0;
struct node
{
    int value, sum;
};
node arr[maxn];
//提取最低位；
int lowbit(int value)
{
    return value&(-value);
}
//单点更新（给value位置上加1）n为所有元素的个数；
void build(int value)
{
    for(int i=value; i<=n; i+=lowbit(i))
    {
        tree[i]+=1;
    }
}
//区间查询,计算数组中1-value的和；
int getsum(int value)
{
    int result = 0;
    for(int i=value; i>=1; i-=lowbit(i))
    {
        result+=tree[i];
    }
    return result;
}
//按value升序排序；
bool cmp(node n1, node n2)
{
    return n1.value<n2.value;
}
int main()
{
    while(scanf("%d",&n)==1&&n)
    {
        ans = 0;
        memset(tree, 0, sizeof(tree));
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &arr[i].value);
            arr[i].sum = i;

        }
        sort(arr+1, arr+n+1, cmp);
        //标记原序列中
        for(int i=1; i<=n; i++)
        {
            num[arr[i].sum] = i;//原序列的第sum个是重新排序后的第i个；
        }
        //
        for(int i=1; i<=n; i++)
        {
            build(num[i]);
            ans+=i-getsum(num[i]);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}