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 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 Stone i+1i+1 or Stone i+2i+2. 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≤hi≤1041≤hi≤104

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Input

Input is given from Standard Input in the following format:

NN
h1h1 h2h2 …… hNhN

Output

Print the minimum possible total cost incurred.

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Sample Input 1 Copy

Copy

4
10 30 40 20

Sample Output 1 Copy

Copy

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

Copy

2
10 10

Sample Output 2 Copy

Copy

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

Copy

6
30 10 60 10 60 50

Sample Output 3 Copy

Copy

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.

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=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll n;
ll dp[maxn];
ll a[maxn];
int main()
{
    gbtb;
    cin>>n;
    repd(i,,n)
    {
        cin>>a[i];
    }
    dp[]=;
    dp[]=;
    dp[]=abs(a[]-a[]);
    repd(i,,n)
    {
        dp[i]=min(dp[i-]+abs(a[i]-a[i-]),dp[i-]+abs(a[i]-a[i-]));

    }
    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 - '';
        }
    }
}