【归并排序+逆序数】poj-2299 Ultra-QuickSort

题目描述

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.

输入

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.

输出

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.

样例输入

5
9
1
0
5
4
3
1
2
3
0

样例输出

6
0


就是让你从小到大排序，每次只能交换相邻元素，求最小次数

#include <bits/stdc++.h>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int M=500005;
int L[M/2+2],R[M/2+2];
int t[M];
ll ans;
int n;
void mergearr(int A[],int left,int mid,int right)
{
    int i=left,j=mid+1;
    int k=left;
    while(i<=mid&&j<=right)
    {
        if(A[i]<=A[j])
            t[k++]=A[i++];
        else{
            t[k++]=A[j++];
            ans+=mid-i+1;
        }
    }
    while(i<=mid)
        t[k++]=A[i++];
    while(j<=right)
        t[k++]=A[j++];
    for(int i=left;i<=right;i++)
        A[i]=t[i];
}

void mergeSort(int A[],int left,int right)
{
    if(left<right)
    {
        int mid=(left+right)/2;
        mergeSort(A,left,mid);
        mergeSort(A,mid+1,right);
        mergearr(A,left,mid,right);
    }
}
int main()
{
    int A[M];
    int n,i;
    while(scanf("%d",&n)&&n!=0)
    {
        ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&A[i]);
        }
         mergeSort(A,1,n);
        printf("%lld\n",ans);
    }

    return 0;
}