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

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

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題目鏈接：https://atcoder.jp/contests/dp/tasks/dp_a

題意：給你一堆石頭，每一個石頭有一個高度，有一只青蛙站在第一個石頭上，青蛙每一次可以跳1-2個石頭，并且產生起跳高度和落地高度的差的消耗。

問你青蛙跳到第N個石頭，最小需要消耗多少能量？

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思路：

簡單的線性DP， 定義dp[i]的狀態意義為青蛙跳到第i個石頭的時候消耗的最小能量，

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

初始狀態定義： dp[1] = 0 ,? dp[2]=| a[2]-a[1] |

dp[2]一定要預處理，狀態轉移只能從i=3開始，因為第二個石頭只能由第一個石頭跳過去。

不這樣定義會wa的。（親測，23333）

我的AC代碼：

#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';}} }

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轉載于:https://www.cnblogs.com/qieqiemin/p/10247378.html

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