atcoder B - Frog 2 （DP）

B - Frog 2

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $100100$ points

Problem Statement

There are $\mathrm{NN}$ stones, numbered $\mathrm{1,2,\dots ,N1,2,\dots ,N}$. For each $\mathrm{ii}$ ($\mathrm{1\le i\le N1\le i\le N}$), the height of Stone $\mathrm{ii}$ is ${\mathrm{hihi}}^{}$.

There is a frog who is initially on Stone $11$. He will repeat the following action some number of times to reach Stone $\mathrm{NN}$:

• If the frog is currently on Stone $\mathrm{ii}$, jump to one of the following: Stone $\mathrm{i+1,i+2,\dots ,i+Ki+1,i+2,\dots ,i+K}$. Here, a cost of $|hi-hj||hi-hj|$ is incurred, where $\mathrm{jj}$ is the stone to land on.

Find the minimum possible total cost incurred before the frog reaches Stone $\mathrm{NN}$.

Constraints

• All values in input are integers.
• $\mathrm{2\le N\le 1052\le N\le 105}$
• $\mathrm{1\le K\le 1001\le K\le 100}$
• $\mathrm{1\le hi\le 1041\le hi\le 104}$

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$ $\mathrm{KK}$
${\mathrm{h1h1}}^{}$ ${\mathrm{h2h2}}^{}$ $\dots \dots$ ${\mathrm{hNhN}}^{}$

Output

Print the minimum possible total cost incurred.

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Sample Input 1 Copy

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5 3
10 30 40 50 20

Sample Output 1 Copy

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30

If we follow the path $11$ → $22$ → $55$, the total cost incurred would be $|10-30|+|30-20|=30|10-30|+|30-20|=30$.

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Sample Input 2 Copy

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3 1
10 20 10

Sample Output 2 Copy

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20

If we follow the path $11$ → $22$ → $33$, the total cost incurred would be $|10-20|+|20-10|=20|10-20|+|20-10|=20$.

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Sample Input 3 Copy

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2 100
10 10

Sample Output 3 Copy

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0

If we follow the path $11$ → $22$, the total cost incurred would be $|10-10|=0|10-10|=0$.

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Sample Input 4 Copy

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10 4
40 10 20 70 80 10 20 70 80 60

Sample Output 4 Copy

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40

If we follow the path $11$ → $44$ → $88$ → $1010$, the total cost incurred would be $|40-70|+|70-70|+|70-60|=40|40-70|+|70-70|+|70-60|=40$.

题目链接：https://atcoder.jp/contests/dp/tasks/dp_b

题意：这一篇的进阶版。

把每次只能跳1~2步改成1~k步。

做法只需要其他的不变，把转移方程那里循环1~k次就行了。

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll n;
ll dp[maxn];
ll a[maxn];
ll k;
int main()
{
    gbtb;
    cin>>n>>k;
    repd(i,1,n)
    {
        cin>>a[i];
    }
    dp[1]=0;
    dp[0]=0;
    repd(i,2,n)
    {
        dp[i]=inf;
    }
    repd(i,2,n)
    {
        for(int j=i-1;j>=max(1ll,i-k);j--)
        {
            dp[i]=min(dp[i],dp[j]+abs(a[i]-a[j]));
        }
    }
    cout<<dp[n];
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}