2014 Multi-University Training Contest 5 HDOJ 4911 Inversion

題目：

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 73    Accepted Submission(s): 26

Problem Description

bobo has a sequence a₁,a₂,…,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≤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

解題思路：

求出序列的逆序數，減去k即爲所求結果。這裏用O（n2）的算法會超時，我們只能用O(nlogn)的算法，可以用樹狀數組離散化求解，也可以用歸併排序求解，ps：逆序數可能會超int範圍。

代碼：

#include<iostream>
#include<fstream>
using namespace std;
const int N=500005;
int a[N];
int t[N];
long long sum;
void copy(int *dest,int *src,int b,int e)
{
	while(b<=e)
	{
		dest[b]=src[b];
		b++;
	}
}

void merge(int * a,int b,int m, int e)
{
	int i=b;
	int j=m+1;
	int k=b;
	while((i<=m)&&(j<=e))
	{
		if(a[i]<=a[j])
			t[k++]=a[i++];

		else
		{
			t[k++]=a[j++];
			sum+=(m-i+1);
		}
	}
	while(i<=m)
	{
		t[k++]=a[i++];
	}
	while(j<=e)
	{
		t[k++]=a[j++];
	}
	copy(a,t,b,e);
}

void mergesort(int * a,int i,int j)
{
		if(i>=j)return;
		int m=(i+j)/2;
		mergesort(a,i,m);
		mergesort(a,m+1,j);
		merge(a,i,m,j);
}

int main()
{
	int n, k;
	std::ios::sync_with_stdio(false);
	while(cin>>n>>k)
	{
		sum=0;
		for(int i=0;i<n;i++)
			cin>>a[i];
		mergesort(a,0,n-1);
		sum -= k;
		if(sum < 0) sum = 0;
		cout<< sum <<endl;
	}
	return 0;
}