• HDU 4911 Inversion （逆序数 归并排序）

    Inversion

    题目链接：

    http://acm.hust.edu.cn/vjudge/contest/121349#problem/A

    Description

    bobo has a sequence a 1,a 2,…,a n. 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 a i>a j.

    Input

    The input consists of several tests. For each tests:

    The first line contains 2 integers n,k (1≤n≤10 5,0≤k≤10 9). The second line contains n integers a 1,a 2,…,a n (0≤a i≤10 9).

    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


    ##题意： 对有n个元素的数组进行不超过k次操作，每次操作允许交换相邻元素; 求操作后能得到的整个数组的最小逆序数.
    ##题解： 容易得出每次交换相邻元素最多只能使得逆序数降低1. 而对于一个逆序数不为0的数组，一定存在相邻的两个数满足(i < j, ai < aj)(可反证). 所以每次操作都可以使得逆序数减少1. 最终答案即为max(0, 逆序数). 求逆序数可以用归并排序、树状数组和线段树(数的规模较小时).
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 101000 #define mod 100000007 #define inf 0x3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std;

    LL num[maxn];
    LL _count=0;

    void merge_array(int left,int mid,int right)
    {
    int temp[right-left+1];int k=0;
    int i=left;int j=mid+1;
    while(i<=mid&&j<=right)
    {
    if(num[i]<=num[j])
    temp[k++]=num[i++];
    else
    {temp[k++]=num[j++];_count+=mid-i+1;}
    }
    while(i<=mid)
    temp[k++]=num[i++];
    while(j<=right)
    temp[k++]=num[j++];

    for(i=left;i<=right;i++)
    num[i]=temp[i-left];

    }

    void merge_sort(int left,int right)
    {
    if(left<right)
    {
    int mid=left+(right-left)/2;
    merge_sort(left,mid);
    merge_sort(mid+1,right);
    merge_array(left,mid,right);
    }
    }

    int main(void)
    {
    //IN;

    int n;
LL k;
while(scanf("%d %I64d",&n,&k) != EOF)
{
    for(int i=0; i<n; i++)
        scanf("%I64d",&num[i]);
    _count=0;
    merge_sort(0, n-1);

    printf("%I64d
", max(_count-k,0LL));
}
return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5719053.html
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