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  • 解题报告——POJ 2299

    Ultra-QuickSort
    Time Limit: 7000MS   Memory Limit: 65536K
    Total Submissions: 44671   Accepted: 16240

    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
    ----------------------------------------------------------------------------------------------------------------
    题目是求逆序对,可以通过快排或者归并算法得出逆序数然后输出。
    ------------------------------------------------------
    #include <stdio.h>
    #include <stdlib.h>
    
    __int64 sum = 0;
    
    void mergeSort(int arr[], int first, int last, int temp[])
    {
        if (first < last)
        {
            int mid = (first + last) / 2;
            mergeSort(arr, first, mid, temp);
            mergeSort(arr, mid + 1, last, temp);
            mergeArray(arr, first, mid, last, temp);
    
        }
    }
    void mergeArray(int arr[], int first, int mid, int last, int temp[])
    {
        int i = first, j = mid + 1;
        int m = mid, n = last;
        int k = 0;
        while (i <= m && j <= n)
        {
            if (arr[i] < arr[j])
            {
                temp[k++] = arr[i++];
            }
            else
            {
                temp[k++] = arr[j++];
                sum += m - i + 1;
            }
        }
    
        while (i <= m)
        {
            temp[k++] = arr[i++];
        }
        while (j <= n)
        {
            temp[k++] = arr[j++];
        }
        for (i = 0; i < k; i++)
        {
            arr[first + i] = temp[i];
        }
    }
    
    void MergeSort(int arr[], int n)
    {
        int *p = (int *)malloc(500000);
        if (p)
        {
            mergeSort(arr, 0, n - 1, p);
            //free(p);
        }
    }
    
    
    int main()
    {
        int head;
        int a[500000];
        //freopen("D:\test.txt", "r", stdin);
        while(1)
        {
            scanf("%d", &head);
            if (head == 0) break;
            int i;
            for (i = 0; i < head; i++)
            {
                scanf("%d", &a[i]);
            }
            sum = 0;
            MergeSort(a, head);
            printf("%I64d
    ", sum);
    
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/codingpenguin/p/4286585.html
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