CF360B Levko and Array （二分查找+DP）

链接：CF360B

题目：

B. Levko and Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Levko has an array that consists of integers: a₁, a₂, ... , a_n. 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:

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, ... , a_n ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

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

Sample test(s)

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.

大意：在数组a中改变任意k个元素的值，使得到的的值最小。

题解：

二分答案，用dp检查可不可行。

dp[j]表示a[0]~a[j]中，a[j]不变时所要改变的数的数量。

 bool check(ll x)
 {
     int i,j;
     for(i=; i<n; i++)
         dp[i]=i;
     for(i=; i<n; i++)
         for(j=i+; j<n; j++)
             if(abs(a[i]-a[j])<=(j-i)*x) dp[j]=min(dp[j],dp[i]+j-i-);
     for(i=; i<n; i++)
         if (dp[i]+n-i- <= k)
             return true;
     return false;
 }

check如上，转移方程时假设a[i]和a[j]之间的数都要改变，那个if是判断是否全部改变了，达成一个超碉的阶梯状，a[i]和a[j]还是差的太远的话就不转移了。

然后后面算的dp[i]+n-i-1是a[i]不变，后面的元素全变的需要变的元素数量。只要有一种方案行，就行。

代码：

 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 #define ll __int64

 int a[];
 int dp[];
 int n,k;
 bool check(ll x)
 {
     int i,j;
     for(i=; i<n; i++)
         dp[i]=i;
     for(i=; i<n; i++)
         for(j=i+; j<n; j++)
             if(abs(a[i]-a[j])<=(j-i)*x) dp[j]=min(dp[j],dp[i]+j-i-);
     for(i=; i<n; i++)
         if (dp[i]+n-i- <= k)
             return true;
     return false;
 }

 int main()
 {
     int i;
     ll mid,l,r;
     while(scanf("%d%d",&n,&k)!=EOF)
     {
         for(i=; i<n; i++)
             scanf("%d",&a[i]);
         l=;
         r=;
         while(l<=r)
         {
             mid=(l+r)>>;
             //cout<<l<<'<'<<mid<<'<'<<r<<endl;
             if(check(mid)) r=mid-;
             else l=mid+;
         }
         printf("%I64d\n",l);
     }
     return ;
 }