POJ 2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 32626		Accepted: 11642

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

给一组数，求这组数的逆序数
开始直接用了线性代数里讲的方法，自己估算了一下O(n^2)，感觉要跪，试着交了一下，果然跪了
百度后发现此题可以用归并排序做，每次merge的时候，当从后半部分取数时，就把当前前半部分剩余数的个数加到ans中去
这题给我最大的启示就是，每个排序算法都有其奇妙的用处，不能只会快排

#include<iostream>

using namespace std;

long long ans;
long a[500000],b[500000];

void Merge(long low,long mid,long high)
{
    long i=low,j=mid+1,k=low,end_1=mid,end_2=high;
    while(i<=end_1&&j<=end_2)
    {
        b[k++]=(a[i]<=a[j]?a[i++]:(ans+=mid-i+1,a[j++]));
    }
    while(i<=end_1)
        b[k++]=a[i++];
    while(j<=end_2)
        b[k++]=a[j++];
    for(i=low;i<=high;i++)
        a[i]=b[i];
}

void Sort(long p,long q)
{
    if(p==q)
        return;
    long mid=(p+q)/2;
    Sort(p,mid);
    Sort(mid+1,q);
    Merge(p,mid,q);
}

int main()
{
    long n;

    while((cin>>n)&&n)
    {
        for(int i=0;i<n;i++)
            cin>>a[i];
        ans=0;
        Sort(0,n-1);
        cout<<ans<<endl;
    }

    return 0;
}