【Atcoder Dp B Frog2】

最新推荐文章于 2024-11-22 21:17:14 发布

ljq真的不能再吃了

最新推荐文章于 2024-11-22 21:17:14 发布

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分类专栏： ACM DP

本文链接：https://blog.youkuaiyun.com/heucodesong/article/details/89367508

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本文详细探讨了Atcoder上的Frog2问题，这是一个与Frog1相似但增加了k步要求的动态规划挑战。在解决过程中，作者通过在长度循环中枚举变量j，更新动态规划状态dp[i]，使其等于dp[i]和dp[i-j]+abs(arr[i]-arr[i-j])的较小值，以此来优化路径选择。

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Atc Frog2
和第一题Frog1其实是一样的只不过有k步要求
那么你在长度循环的时候枚举一个j
dp[i] = min(dp[i],dp[i-j]+abs(arr[i]-arr[i-j]));

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 100025;
ll dp[MAX_N],arr[MAX_N];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,k;scanf("%d%d",&n,&k);
    for(int i = 1;i<=n;++i) scanf("%lld",&arr[i]);
    for(int i = 2;i<=n;++i)
    {
        dp[i] = 0x3f3f3f3f;
        for(int j = 1;j<=k;++j)
        {
            if(i-j<1) break;
            dp[i] = min(dp[i],dp[i-j]+abs(arr[i]-arr[i-j]));
        }
    }
    printf("%lld\n",dp[n]);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}