• hdu 4911 Inversion


    Inversion

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 197    Accepted Submission(s): 82


    Problem Description
    bobo has a sequence a1,a2,…,an. He is allowed to swap two adjacent numbers for no more than k times.

    Find the minimum number of inversions after his swaps.

    Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and ai>aj.
     

    Input
    The input consists of several tests. For each tests:

    The first line contains 2 integers n,k (1≤n≤105,0≤k≤109). The second line contains n integers a1,a2,…,an (0≤ai≤109).
     

    Output
    For each tests:

    A single integer denotes the minimum number of inversions.
     

    Sample Input
    3 1 2 2 1 3 0 2 2 1
     

    Sample Output
    1 2
     

    Author
    Xiaoxu Guo (ftiasch)
     

    Source
     



    题解及代码:


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    __int64 ans=0;
    
    void merge(int Array[],int l,int m,int r)
    {
        int temp[100110];
        int i=l,j=m+1,k=0;
        while(i<=m&&j<=r)
        {
            if(Array[i]<=Array[j])
            {
                temp[k++]=Array[i];
                i++;
            }
            else
            {
                temp[k++]=Array[j];
                j++;
                ans+=(m-i+1);     //求逆序数的关键
            }
        }
        while(i<=m) {temp[k++]=Array[i];i++;}
        while(j<=r) {temp[k++]=Array[j];j++;}
    
        for(i=l,j=0;i<=r&&j<k;i++,j++)
        {
            Array[i]=temp[j];
        }
    }
    
    void mergesort(int Array[],int l,int r)
    {
        if(l>=r) return;
        int m=(l+r)/2;
    
        mergesort(Array,l,m);
        mergesort(Array,m+1,r);
    
        merge(Array,l,m,r);
    }
    
    int main()
    {
        int n,k;
        int Array[100110];
        while(scanf("%d%d",&n,&k)!=EOF)
        {
            ans=0;
            for(int i=0;i<n;i++)
            {
                cin>>Array[i];
            }
            mergesort(Array,0,n-1);
            if(k>=ans)
            {
                printf("0
    ");
            }
            else printf("%I64d
    ",ans-k);
        }
        return 0;
    }
    /*
    简单题,本题主要是求出当前数列的逆序数。
    
    我们直到每次调序都能将逆序数降低1。我们先求出逆序数。
    然后推断,当给出的调序次数小于当前序列的逆序数时,输出ans-k,
    否则输出0。

    */



    版权声明:本文博主原创文章。博客,未经同意不得转载。

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  • 原文地址:https://www.cnblogs.com/lcchuguo/p/4826044.html
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