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  • POJ 2299 归并排序模板

    Ultra-QuickSort
    Time Limit: 7000MS   Memory Limit: 65536K
    Total Submissions: 49222   Accepted: 18021

    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
    

    Source

     
     
     
     
    题意:相邻位置的两个数交换,多少次可以把数组变成从小到大的顺序!
    归并排序的思想(线代学的逆序数)!
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #define N 500010
    #define LL long long
    using namespace std;
    LL a[N], b[N], sum;
    void Merge(int left, int mid, int right)
    {
        int i = left, j = mid + 1;
        int k = 0;
        while(i <= mid && j <= right)
        {
            if(a[i] <= a[j])
                b[k++] = a[i++];
            else
            {
                b[k++] = a[j++];
                sum += mid - i + 1;
            }
        }
        while(i <= mid)
            b[k++] = a[i++];
        while(j <= right)
            b[k++] = a[j++];
        for(i = 0; i < k; i++)
            a[left + i] = b[i];
    }
    void Mysort(int left, int right)
    {
        if(left < right)
        {
            int mid = (right + left) / 2;
            Mysort(left, mid);
            Mysort(mid + 1, right);
            Merge(left, mid, right);
        }
    }
    int main()
    {
        int n, i;
        while(scanf("%d", &n), n)
        {
            sum = 0;
            memset(a, 0, sizeof(a));
            for(i = 0; i < n; i++)
                scanf("%lld", &a[i]);
            Mysort(0, n - 1);
            printf("%lld
    ", sum);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/yu0111/p/4792010.html
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