POJ 2299 Ultra-QuickSort （归并排序）

Ultra-QuickSort

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

算法课上讲过这个，正好学习下逆序对的求法。

#include<iostream>
#include<stdio.h>
#define M 500010
int a[M];
int aux[M];
long long int  ans;
using namespace std;
void merge(int a[],int l,int mid,int h)
{
    for(int k=l;k<=h;++k)
        aux[k]=a[k];
    int i=l;
    int j=mid+1;
    for(int k=l;k<=h;++k)
    {
        if(i>mid)a[k]=aux[j++];
        else if(j>h)a[k]=aux[i++];
        else if(aux[i]<aux[j])a[k]=aux[i++];
        else {
            a[k]=aux[j++];
            ans+=mid-i+1;//左边的元素比右边的大，那么左边剩下的没比完的元素和这个右边的元素都构成了逆序对！
        }
    }
}
void sort(int a[],int l,int h)
{
    if(l>=h)return ;
    int mid=l+(h-l)/2;
    sort(a,l,mid);
    sort(a,mid+1,h);
    merge(a,l,mid,h);
}
int main(int argc, char *argv[])
{
    //freopen("2299.in","r",stdin);
    int n;
    while(scanf("%d",&n)&&n!=0){
        ans=0;
        for(int i=0;i<n;++i)
            scanf("%d",&a[i]);
        sort(a,0,n-1);
        printf("%lld\n",ans);
    }
    return 0;
}