codeforces D. Levko and Array（二分加dp）挺好的一个题

最新推荐文章于 2022-02-13 23:54:15 发布

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最新推荐文章于 2022-02-13 23:54:15 发布

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分类专栏：动态规划二分文章标签： codeforces dp

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Levko has an array that consists of integers: a1, a2, ... , an. But he doesn’t like this array at all.

Levko thinks that the beauty of the array a directly depends on value c(a), which can be calculated by the formula:

$\text{[math]}$ The less value c(a) is, the more beautiful the array is.

It’s time to change the world and Levko is going to change his array for the better. To be exact, Levko wants to change the values of at most k array elements (it is allowed to replace the values by any integers). Of course, the changes should make the array as beautiful as possible.

Help Levko and calculate what minimum number c(a) he can reach.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2000). The second line contains space-separated integers a1, a2, ... , an( - 109 ≤ ai ≤ 109).

Output

A single number — the minimum value of c(a) Levko can get.

        Examples 
      
          input 
        
5 2
4 7 4 7 4

          output 
        
0

          input 
        
3 1
-100 0 100

          output 
        
100

          input 
        
6 3
1 2 3 7 8 9

          output 
        
1

Note

In the first sample Levko can change the second and fourth elements and get array: 4, 4, 4, 4, 4.

In the third sample he can get array: 1, 2, 3, 4, 5, 6.

题意：给你n个数你最多改变其中的k个数使得两个相邻的数之间的差的绝对值最小。

思路：二分答案（二分任意差的绝对值看在mid的情况下是否符合情况如果符合情况那么r=mid-1 否则l=mid+1 ）

而在判断的时候就要用到dp dp[i] 表示在不改动a[i] 的情况下前i个数需要改动的个数。

首先dp[1]=0; 那么dp[i] = ? 当然取决于前边的情况如果到j需要改动dp[j] 个数那么对于i 就有dp[i]=dp[j]+ i-j-1 ; 表示（j，i）的数都需要改变（开区间）而改变的条件就是a[i]和

a[j] 的差值小于等于（i-j）*mid 而是否符合条件只需要看dp[i]+n-i 是否小于等于k

d代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define N 2005

using namespace std;

typedef long long ll;

const ll inf =1e15;

int n,k;
ll a[N];
ll dp[N];


int jud(ll num)
{
	memset(dp,inf,sizeof(dp));
	
	dp[1]=0;
	for(int i=2;i<=n;i++)
	{
		dp[i]=i-1;
		for(int j=1;j<i;j++)
		{
			if(abs(a[i]-a[j])<=(i-j)*num)
			{
				dp[i]=min(dp[i],dp[j]+i-j-1);
			}
		}
		if(dp[i]+n-i<=k) return 1;
	}
	
	return 0;
}

int main()
{
	cin>>n>>k;
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	
	ll l ,r,mid;
	l=0; r=inf;
	ll ans=0;
	while(l<=r)
	{
		mid=(l+r)>>1;
		//printf(" mid : %lld\n",mid);
		if(jud(mid))
		{
			ans=mid;
			r=mid-1;
		}
		else l=mid+1;
	}
	
	printf("%lld\n",ans);
	return 0;
} 

/*

20 17
-5 -9 11 -7 -17 -8 0 -14 -20 -15 7 -13 0 -3 -14 0 9 -10 6 -19

5 1
-1000000000 1000000000 -1000000000 1000000000 -1000000000


*/