atcoder A - Frog 1（DP）

A - Frog 1

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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 Stone $\mathrm{i+1i+1}$ or Stone $\mathrm{i+2i+2}$. 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.
• $\mathrm{2\le N\le 1052\le N\le 105}$
• $\mathrm{1\le hi\le 1041\le hi\le 104}$

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Input

Input is given from Standard Input in the following format:

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

Output

Print the minimum possible total cost incurred.

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Sample Input 1 Copy

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4
10 30 40 20

Sample Output 1 Copy

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30

If we follow the path $11$ → $22$ → $44$, 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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2
10 10

Sample Output 2 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 3 Copy

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6
30 10 60 10 60 50

Sample Output 3 Copy

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40

If we follow the path $11$ → $33$ → $55$ → $66$, the total cost incurred would be $|30-60|+|60-60|+|60-50|=40|30-60|+|60-60|+|60-50|=40$.

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

题意：给你一堆石头，每一个石头有一个高度，有一只青蛙站在第一个石头上，青蛙每一次可以跳1-2个石头，并且产生起跳高度和落地高度的差的消耗。

问你青蛙跳到第N个石头，最小需要消耗多少能量？

思路：

简单的线性DP， 定义dp[i]的状态意义为青蛙跳到第i个石头的时候消耗的最小能量，

转移方程即为：dp[i]=min(dp[i-2]+abs(a[i]-a[i-2]),dp[i-1]+abs(a[i]-a[i-1]))

初始状态定义： dp[1] = 0 ,  dp[2]=| a[2]-a[1] |

dp[2]一定要预处理，状态转移只能从i=3开始，因为第二个石头只能由第一个石头跳过去。

不这样定义会wa的。（亲测，23333）

我的AC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];
int main()
{
    gbtb;
    cin>>n;
    repd(i,1,n)
    {
        cin>>a[i];
    }
    dp[1]=0;
    dp[0]=0;
    dp[2]=abs(a[2]-a[1]);
    repd(i,3,n)
    {
        dp[i]=min(dp[i-2]+abs(a[i]-a[i-2]),dp[i-1]+abs(a[i]-a[i-1]));

    }
    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';
        }
    }
}