Inversion

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 528 Accepted Submission(s): 228

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

Sample Output

1
2

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

解法：求的所给序列的逆序数。然后减掉k，假设小于零就去零！

代码例如以下：（归并排序法）

#include<stdio.h>

int is1[112345],is2[112345];// is1为原数组，is2为暂时数组。n为个人定义的长度

__int64 merge(int low,int mid,int high)
{
	int i=low,j=mid+1,k=low;
	__int64 count=0;
	while(i<=mid&&j<=high)
		if(is1[i]<=is1[j])// 此处为稳定排序的关键，不能用小于
			is2[k++]=is1[i++];
		else
		{
			is2[k++]=is1[j++];
			count+=j-k;// 每当后段的数组元素提前时。记录提前的距离
		}
		while(i<=mid)
			is2[k++]=is1[i++];
		while(j<=high)
			is2[k++]=is1[j++];
		for(i=low;i<=high;i++)// 写回原数组
			is1[i]=is2[i];
		return count;
}
__int64 mergeSort(int a,int b)// 下标，比如数组int is[5]，所有排序的调用为mergeSort(0,4)
{
	if(a<b)
	{
		int mid=(a+b)/2;
		__int64 count=0;
		count+=mergeSort(a,mid);
		count+=mergeSort(mid+1,b);
		count+=merge(a,mid,b);
		return count;
	}
	return 0;
}
int main()
{
	int n, x;
	__int64 k;
	__int64 sum;
	while(scanf("%d%I64d",&n,&k)!=EOF)
	{
		for(int i=0;i<n;i++)
		{
			scanf("%d",&x);
			is1[i] = x;
		}
		__int64 ans=mergeSort(0,n-1);
		sum=0;
		printf("%I64d
",ans-k>0?ans-k:0); 
	}
	return 0;
}