题源

源石虫
（小声哔哔：这题当时脑子抽了死活没想到是DP，一直用贪心试，拿来凑个DP专题的数）

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define For for (ll i = 1; i <= n; i++)
#define rFor for (ll i = n; i > 0; i--)
#define rep(i, sta, end) for (ll i = sta; i <= end; i++)
#define rrep(i, end, sta) for (ll i = end; i >= sta; i--)
#define ALL(x) for (auto item : x)inline void solve() {ll n,a,b;cin>>n>>a>>b;vector<ll> w(n+10),cla(n+10);vector<ll > bug1(n+10),bug2(n+10);For{cin>>w[i];}For{cin>>cla[i];}For{if(cla[i]==1){bug1[i]=max(bug1[i-1],bug1[max(i-1-a,0LL)]+w[i]);bug2[i]=bug2[i-1];}else {bug1[i]=bug1[i-1];bug2[i]=max(bug2[i-1],bug2[max(0LL,i-1-b)]+w[i]);}}cout<<bug1[n]+bug2[n]<<endl;
}
int main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);ll num = 1;//cin >> num;while (num--) {//cout << "Main function execute properly before solve function " << endl;solve();//	cout << "Main function execute properly after solve function" << endl;}
}

KEY part解析

For{if(cla[i]==1){bug1[i]=max(bug1[i-1],bug1[max(i-1-a,0LL)]+w[i]);bug2[i]=bug2[i-1];}else {bug1[i]=bug1[i-1];bug2[i]=max(bug2[i-1],bug2[max(0LL,i-1-b)]+w[i]);}}

主要矛盾点在于如何解决杀死了这个虫子就不能杀那一个之间的矛盾问题，表现在DP上的思想主要是，比较当前的虫子不杀死（直接延续上一个位置所能获得的最大金币）和杀死这个虫子（追溯到最近能杀死的与他类型相同的虫子的位置所能获得的最大金币）之间两种选择哪一个爆的金币更多。