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  • 【归并排序+逆序数】poj-2299 Ultra-QuickSort

    题目描述

    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.

    输入

    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.

    输出

    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.

    样例输入

    5
    9
    1
    0
    5
    4
    3
    1
    2
    3
    0
    

    样例输出

    6
    0


    就是让你从小到大排序,每次只能交换相邻元素,求最小次数
    #include <bits/stdc++.h>
    #define ll long long
    #define INF 0x3f3f3f3f
    using namespace std;
    const int M=500005;
    int L[M/2+2],R[M/2+2];
    int t[M];
    ll ans;
    int n;
    void mergearr(int A[],int left,int mid,int right)
    {
        int i=left,j=mid+1;
        int k=left;
        while(i<=mid&&j<=right)
        {
            if(A[i]<=A[j])
                t[k++]=A[i++];
            else{
                t[k++]=A[j++];
                ans+=mid-i+1;
            }
        }
        while(i<=mid)
            t[k++]=A[i++];
        while(j<=right)
            t[k++]=A[j++];
        for(int i=left;i<=right;i++)
            A[i]=t[i];
    }
    
    void mergeSort(int A[],int left,int right)
    {
        if(left<right)
        {
            int mid=(left+right)/2;
            mergeSort(A,left,mid);
            mergeSort(A,mid+1,right);
            mergearr(A,left,mid,right);
        }
    }
    int main()
    {
        int A[M];
        int n,i;
        while(scanf("%d",&n)&&n!=0)
        {
            ans=0;
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&A[i]);
            }
             mergeSort(A,1,n);
            printf("%lld
    ",ans);
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Diliiiii/p/9389845.html
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