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  • poj 2299 Ultra-QuickSort

                                                  http://poj.org/problem?id=2299                     
                                                                     Ultra-QuickSort
    Time Limit: 7000MS   Memory Limit: 65536K
    Total Submissions: 40977   Accepted: 14817

    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

    题意就是求逆对数。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    __int64 a[5000005],c[5000005];
    __int64 k;
    void px(int low,int mid,int hight)
    {
        int i=low,cnt=0,j=mid+1;
        while(i<=mid&&j<=hight)
         {
            if(a[i]>=a[j])
                  {
                     c[cnt++]=a[j++];
                      k+=mid-i+1;
                  }
             else
               c[cnt++]=a[i++];
         }
         while(i<=mid)
         {
             c[cnt++]=a[i++];
         }
         while(j<=hight)
         {
             c[cnt++]=a[j++];
         }
          i=low;cnt=0;
         while(i<=hight)
         {
             a[i++]=c[cnt++];
         }
    }
    void gb(int low,int hight)
    {
        int mid=(low+hight)/2;
        if(low<hight)
        {
        gb(low,mid);
        gb(mid+1,hight);
        px(low,mid,hight);
        }
    }
    int main()
    {
         int i,n;
        while(~scanf("%d",&n)&&n!=0)
        {
            k=0;
          for(i=0;i<n;i++)
           scanf("%I64d",&a[i]);
           gb(0,n-1);
           printf("%I64d
    ",k);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cancangood/p/3893969.html
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