POJ-2299Ultra-QuickSort(BIT|归并排序求逆序对)

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#归并排序 #poj

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本文介绍了一种名为Ultra-QuickSort的排序算法，并通过代码实现分析了该算法如何计算输入序列所需的最少交换次数来达到升序排列。文章还提供了一个具体的示例，展示了算法的工作流程。

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Ultra-QuickSort
Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 62659 Accepted: 23342
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

Waterloo local 2005.02.05

题意：

求逆序对。。。
经典问题，利用归并排序，树状数组等都可以解决
用归并就是要注意有相同元素的时候，a[R]和a[L]相同时应该让a[L]先出去

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long LL;
const int maxn=5e5+5;

LL ans;
int a[maxn],n,b[maxn];

inline void merge(int l,int r)
{
  if(l>=r)return;
  int mid=(l+r)>>1;
  merge(l,mid);
  merge(mid+1,r);
  int L=l,R=mid+1;
  int cnt=l;
  while(L<=mid||R<=r)
  {
    if(L>mid||(a[R]<a[L]&&R<=r))
    {
      b[cnt++]=a[R++];
      ans+=mid-L+1;
    }
    else b[cnt++]=a[L++];
  }
  for(int i=l;i<=r;++i)a[i]=b[i];
}

int main()
{
  while(1)
  {
    scanf("%d",&n);
    if(n==0)break;
    for(int i=1;i<=n;++i)scanf("%d",&a[i]); 
    ans=0;
    merge(1,n);
    printf("%lld\n",ans);
  }
  return 0;
}