A - Frog 1

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 $|30-60|+|60-60|+|60-50|=40|30-60|+|60-60|+|60-50|=40$.

 

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod =1e9+7;
const int N = 1e5+5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
bool prime(int x) {
    if (x < 2) return false;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}      

int main()
{
    int n;
    cin >> n;
    vector<int> a(n + 1);
    rep(i, 1, n)
        cin >> a[i];
    vector<int> dp(n + 1,0);
    a[0] = a[1];
    rep(i, 2, n)
    {
        dp[i] = min(dp[i-1]+abs(a[i] - a[i - 1]),dp[i-2] + abs(a[i] - a[i - 2]));
    }
    cout << dp[n] << endl;
    return 0;
}