poj 2299 Ultra-QuickSort(树状数组求逆序数)

本文介绍了一种使用树状数组求解逆序数的方法,并通过一道具体题目进行实践。文章提供了完整的C++代码实现,包括离散化处理、树状数组的更新与查询操作等关键步骤。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

求逆序数的模板题。

讲用树状数组求逆序数不错的博客:点击打开链接


注意需要离散化,ans需要用long long


代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 5*1e5+5;
int tree[maxn], ra[maxn], n;
struct node
{
    int val, index;
    bool operator < (const node &a) const
    {
        return val < a.val;
    }
}a[maxn];

int lowbit(int i) { return i&(-i); }

void update(int pos, int val)
{
    while(pos <= n)
    {
        tree[pos] += val;
        pos += lowbit(pos);
    }
}

int query(int pos)
{
    int ans = 0;
    while(pos > 0)
    {
        ans += tree[pos];
        pos -= lowbit(pos);
    }
    return ans;
}

int main(void)
{
    while(cin >> n, n)
    {
        memset(tree, 0, sizeof(tree));
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i].val);
            a[i].index = i;
        }
        sort(a+1, a+n+1);
        for(int i = 1; i <= n; i++)
            ra[a[i].index] = i;
        long long ans = 0;
        for(int i = 1; i <= n; i++)
        {
            update(ra[i], 1);
            ans += i-query(ra[i]);
        }
        printf("%lld\n", ans);
    }
    return 0;
}

Ultra-QuickSort
Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 56638 Accepted: 20934

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

Source



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值