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 NN stones, numbered 1,2,…,N1,2,…,N. For each ii (1≤i≤N1≤i≤N), the height of Stone ii is hihi.

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

• If the frog is currently on Stone ii, jump to one of the following: Stone i+1,i+2,…,i+Ki+1,i+2,…,i+K. Here, a cost of |hi−hj||hi−hj| is incurred, where jj is the stone to land on.

Find the minimum possible total cost incurred before the frog reaches Stone NN.

Constraints

• All values in input are integers.
• 2≤N≤1052≤N≤105
• 1≤K≤1001≤K≤100
• 1≤hi≤1041≤hi≤104

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Input

Input is given from Standard Input in the following format:

NN KK
h1h1 h2h2 …… 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=;
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,,n)
    {
        cin>>a[i];
    }
    dp[]=;
    dp[]=;
    repd(i,,n)
    {
        dp[i]=inf;
    }
    repd(i,,n)
    {
        for(int j=i-;j>=max(1ll,i-k);j--)
        {
            dp[i]=min(dp[i],dp[j]+abs(a[i]-a[j]));
        }
    }
    cout<<dp[n];
    return ;
}

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