标签：tco line title 需要 statement ber element stat etc
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 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:
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.
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‘; } } }
标签：tco line title 需要 statement ber element stat etc
原文地址：https://www.cnblogs.com/qieqiemin/p/10247378.html
