poj 2299 Ultra-QuickSort 归并排序解法

点击打开链接题目链接

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

要求求出冒泡排序的交换次数

可以用归并排序求出逆序数然后做

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int MAXN=500010;
__int64 t;
int num[MAXN],R[MAXN];
void merco(int l,int mid,int r)
{
    int i=l,j=mid+1,p=0;
    while(i<=mid&&j<=r)
    {
        if(R[i]<=R[j])
        {
            num[p++]=R[i++];
        }
        else
        {
            num[p++]=R[j++];
            t+=(mid-i+1);
        }
    }
    while(i<=mid)
    {
        num[p++]=R[i++];
    }
    while(j<=r)
    {
        num[p++]=R[j++];
    }
    for(i=0;i<p;i++)
    {
        R[l+i]=num[i];
    }
}
void mer(int l,int r)
{
    int mid=(l+r)>>1;
    if(l<r)
    {
        mer(l,mid);
        mer(mid+1,r);
        merco(l,mid,r);
    }
}
int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF&&n)
    {
        t=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&R[i]);
        }
        mer(0,n-1);
        printf("%I64d\n",t);
    }
    return 0;
}