• 线段树——Ultra-QuickSort


    题目网址:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=109331#problem/A

    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:输入n个数,相邻两个数可以交换位置进行从小到大排序,求最小交换次数。

    解题思路:用线段树的思想将数组一半一半的划分成小的区间,最后划分为只有两个数的区间,这两个数进行比较交换位置,记录交换次数就。先进行前两个数的比较,然后扩大区间为(left,right),
    可知(left,right)区间中,(left,center)和(center+1,right)区间里的数已经排好序,将(center+1,right)中的数一个一个与(left,center)中的数比较大小,用另一个数组记录
    新的排序后的位置(相当于向前插入数),并记录交换次数。
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <cstdlib>
    using namespace std;
    int a[500010];
    long long num;
    int temp[500010];
    
    void Merge(int arr[], int left, int center, int right)
    {
        int i=left;
        int j=center+1;
        int k=0;
        while (i<=center&&j<=right)
        {
            ///可知(left,right)区间中,(left,center)和(center+1,right)区间里的数已经排好序
            if (arr[i] > arr[j])
            {
                temp[k++] = arr[j++];///
                num+= center+1-i;
            }
            else
                temp[k++] = arr[i++];///记录小的数,以便得到排序后新的序列;
        }
        while (i <= center) ///
            temp[k++] = arr[i++];
        while (j <= right)  ///这个和上个while语句两个while语句记录了原数组交换次序后的新的数组;
            temp[k++] = arr[j++];
        for (i = left, k = 0; i <= right; i++, k++)///将array[]数组变为新的数组;
            arr[i] = temp[k];
    }
    
    void mergeSort(int arr[], int left, int right)
    {
        if (left<right)
        {
            int center = (left + right) / 2;
            mergeSort(arr, left, center);     ///
            mergeSort(arr, center + 1, right);///将数组划分为两个区间;
            Merge(arr, left, center, right);  ///将这个区间里的数进行排序;
        }
    }
    
    int main()
    {
       int n;
       while(scanf("%d",&n)&&n)
       {
          num=0;
          for(int i=0;i<n;i++)
          scanf("%d",&a[i]);
          mergeSort(a,0,n-1);
          printf("%lld
    ",num);
       }
    }
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  • 原文地址:https://www.cnblogs.com/chen9510/p/5281276.html
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