HDU 4911 Inversion

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1488    Accepted Submission(s): 606

Problem Description

bobo has a sequence a₁,a₂,…,a_n. He is allowed to swap twoadjacent 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≤10⁵,0≤k≤10⁹). The second line contains n integers a₁,a₂,…,a_n (0≤a_i≤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
   
   
   
   

 

Source

2014 Multi-University Training Contest 5


题目链接  :http://acm.hdu.edu.cn/showproblem.php?pid=4911


题目大意  :给出n个数，每次可以交换相邻的两个数，最多交换k次，求交换后最小的逆序数是多少。

题目分析  :若逆序数大于0，则存在1 ≤ i < n，使得交换a[i]和a[i+1]后逆序数减1，所以最后的答案就是max((原数列逆序数对数 - k), 0)。利用归并排序求出原序列的逆序对数即可

#include <cstdio>
#define ll long long
const int MAX = 1e5 + 5;
int a[MAX]; 
ll num;  //逆序对个数

//求逆序对个数模板
void mergg(int *arr,int low, int mid, int high)
{
    int* b = new int[high + 1];      //b[] 复制 arr[]
    for (int i = low;i <= high;++i)
        b[i] = arr[i];
    int i = low, j = mid + 1, k = low;
    while (i <= mid && j <= high)
    {
        if (b[i] <= b[j])
            arr[k++] = b[i++];
        else
        {
            //发生逆序，此时由于a[i..mid]是已经有序了，
            //那么a[i+1], a[i+2], ... a[mid]都是大于a[j]的，
            //都可以和a[j]组成逆序对，因此number += mid - i + 1
            arr[k++] = b[j++];
            num += mid - i + 1;  
        }
    }
    while (i <= mid) 
        arr[k++] = b[i++]; //这两种情况只会发生一种
    while (j <= high)
        arr[k++] = b[j++];    
    delete []b;
}

void merge_sort(int *arr,int low,int high)  //二路归并排序
{
    if (low < high)
    {
        int mid = low + (high - low) / 2; //防溢出
        merge_sort(arr, low, mid);
        merge_sort(arr, mid + 1, high);
        mergg(arr, low, mid, high);
    }
}

int main()
{
    int n, k;
    while(scanf("%d%d",&n,&k) != EOF)
    {    
        num = 0;
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        merge_sort(a, 0, n-1);
        printf("%I64d\n", num - k > 0 ? num - k : 0);
    }
}