Ultra-QuickSort poj2299 （归并排序 求逆序数对）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 57649		Accepted: 21298

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

///2299 MergeSort
#include <iostream>
#include <cstdio>
using namespace std;

const int maxn=500010;

int num[maxn],temp[maxn];
long long sum;

void Merge(int low,int mid,int high)
{
    int i=low;
    int j=mid+1;
    int cnt=1; ///cnt 多加了一个

    while(i<=mid && j<=high)
    {
        if(num[i] < num[j]) ///防止相等的时候计算逆序对
        {
            temp[cnt++]=num[i++];
        }
        else
        {
            temp[cnt++]=num[j++];
            sum+=mid+1-i;
            ///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
            /// 12354  5和4 比较; 逆序对就是1  12543 5和4 然后就是12453 4和3
        }
    }

    while(i<=mid)
    {
        temp[cnt++]=num[i++];
    }

    while(j<=high)
    {
        temp[cnt++]=num[j++];
    }

    for (i=1;i<cnt;i++) /// 从low开始复制回去
    {
        num[low+i-1]=temp[i];
    }
}

void msort(int low,int high)
{
    int mid;
    if (low<high)
    {
        mid=(low+high)/2;

        msort(low,mid);
        msort(mid+1,high);
        Merge(low,mid,high);
    }
}

int main()
{
    int i,n;
    while(cin>>n && n)
    {
    sum=0;
        for(i=1; i<=n; i++)
        {
            scanf("%d",&num[i]);
        }
        msort(1,n);
        printf("%lld\n",sum);
    }
    return 0;
}