poj 2299 Ultra-QuickSort（树状数组+离散化的题目）据说是简单题，不过还是觉得好难。。。

最新推荐文章于 2019-08-02 17:17:52 发布

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本文介绍了一种名为Ultra-QuickSort的排序算法，并详细分析了其工作原理及实现过程。通过交换相邻元素的方式对序列进行排序，文章提供了一个使用树状数组优化的解决方案，以减少大规模数据排序所需的交换次数。

1、http://poj.org/problem?id=2299

2、题目大意：

一个序列有n个数字，他们是无序的，现在要将这些数字交换顺序，使得他们是从小到大排列的，输出最少交换几次，3、思路分析

如果数字个数很小的话，实际上是可以用冒泡排序的，冒泡交换的步数实际上就是所求，但是数据很大，所以不能这么做，看网上解题报告用树状数组解决，作为第一个树状数组的题目，真心觉得好难，还得继续看，

还有一个问题就是除了n非常大不能用冒泡做，还有就是这n个数字也是非常大的，所以要用到离散化

4、题目

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 37365		Accepted: 13438

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

Sample Output

6
0

Source

Waterloo local 2005.02.05

5、AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 500005
int b[N],n,c[N];
struct node
{
    int id;
    int x;
} a[N];
int cmp(node a,node b)
{
    return a.x<b.x;
}
void update(int i,int x)
{
    while(i<=n)
    {
        c[i]+=x;
        i+=i&(-i);
    }
}
int sum(int i)
{
    int tmp=0;
    while(i>0)
    {
        tmp+=c[i];
        i-=i&(-i);
    }
    return tmp;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i].x);
            a[i].id=i;
        }
        sort(a+1,a+n+1,cmp);
        //离散化
        a[0].x=-1;
        for(int i=1; i<=n; i++)
        {
            if(a[i].x!=a[i-1].x)
                b[a[i].id]=i;
            else
                b[a[i].id]=b[a[i-1].id];
        }
        long long ans=0;
        for(int i=1; i<=n; i++)
        {
            //每次更新后判断左边比当前数大的数的个数
            update(b[i],1);
            ans+=sum(n)-sum(b[i]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}