Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 48111		Accepted: 17549

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

题意就是求所给序列的逆序数
可以用归并排序求
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define LL long long
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define WW freopen("output.txt","w",stdout)
#define RR freopen("input.txt","r",stdin)

using namespace std;

const int MAX= 500100 ;

int a[MAX];
int b[MAX];
int n;
LL sum;//开始估计错值，用int 直接爆掉了
void Link_Sqrt(int low,int mid,int Low,int high)
{
    int i=low,j=Low,s=0;
    while(i<=mid&&j<=high)
    {
        while(a[i]<=a[j]&&i<=mid&&j<=high)
        {
            b[s++]=a[i++];
        }
        while(a[j]<a[i]&&i<=mid&&j<=high)//当a[i]>a[j]时，因为a[i--mid]与a[mid=1--high]都是有序序列，所以对与a[j]他的逆序数就是mid-i+1
        {
            sum+=(mid-i+1);
            b[s++]=a[j++];
        }
    }
    while(i<=mid)
    {
        b[s++]=a[i++];
    }
    while(j<=high)
    {
        b[s++]=a[j++];
    }
    for(int k=0;k<s;k++)
    {
        a[low+k]=b[k];
    }
}
void Sqrt(int low,int high)
{
    int i=low,j=high;
    if(i>=j)
    {
        return ;
    }
    int mid=(i+j)/2;//二分
    Sqrt(i,mid);
    Sqrt(mid+1,j);
    Link_Sqrt(i,mid,mid+1,j);//归并
}

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