A - Frog 1
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 100100 points
There are NN stones, numbered 1,2,…,N1,2,…,N. For each ii (1≤i≤N1≤i≤N), the height of Stone ii is
There is a frog who is initially on Stone 11. He will repeat the following action some number of times to reach Stone NN:
Find the minimum possible total cost incurred before the frog reaches Stone NN.
Input is given from Standard Input in the following format:
NN h1h1 h2h2 …… hNhN
Print the minimum possible total cost incurred.
4 10 30 40 20
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.
2 10 10
0
If we follow the path 11 → 22, the total cost incurred would be |10−10|=0|10−10|=0.
6 30 10 60 10 60 50
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.
題目連結：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，有一隻青蛙站在第一個石頭上，青蛙每一次可以跳1-2個石頭，並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？
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程式碼：
#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'; } } }