POJ 2299 Ultra-QuickSort

最新推荐文章于 2020-03-19 20:17:52 发布

Orion_Rigel

最新推荐文章于 2020-03-19 20:17:52 发布

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分类专栏：排序文章标签： poj

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本文介绍了一个特定的排序算法Ultra-QuickSort，该算法通过计算输入序列的逆序对数量来确定排序所需的交换次数。文章提供了一段使用归并排序实现的C++代码示例。

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.

【题目分析】
扯了半天，就是求一个逆序对的个数。归并排序解决。

【代码】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int a[500001];
int b[500001];
long long ans;
int n;
inline void mer(int l,int r)
{
    if (l==r) return ;
    int mid=(l+r)/2;
    mer(l,mid); mer(mid+1,r);
    int t1=l,t2=mid+1,t3=l;
    while (t3<=r)
    {
        if (t1>mid) {b[t3++]=a[t2++]; continue;}
        if (t2>r) {b[t3++]=a[t1++]; continue;}
        if (a[t1]<a[t2]) b[t3++]=a[t1++];
        else b[t3++]=a[t2++],ans+=mid-t1+1;
    }
    for (int i=l;i<=r;++i) a[i]=b[i];
    return ;
}
int main()
{
    while (scanf("%d",&n)==1&&n)
    {
        ans=0;
        for (int i=1;i<=n;++i) scanf("%d",&a[i]);
        mer(1,n);
        cout<<ans<<endl;
    }
}