Help for USACO 2017 January Contest, Gold Problem 2. Hoof, Paper, Scissors

Problems and Contests
#1

For the problem link,
I referred to the solution given below -

#include <bits/stdc++.h> // see /general/running-code-locally
using namespace std;

using ll = long long;

using vi = vector<int>;
#define pb push_back
#define all(x) begin(x), end(x)
#define sz(x) (int) (x).size()

using pi = pair<int,int>;
#define f first
#define s second
#define mp make_pair

void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0); // see /general/fast-io
	if (sz(name)) {
		freopen((name + ".in").c_str(), "r", stdin); // see /general/input-output
		freopen((name + ".out").c_str(), "w", stdout);
	}
}
//EndCodeSnip

const int MX = 1e5+5;

int dp[MX][25][3]; // dp[i][j][k] is the largest number of games she wins at games i with switches j with current item k
int A[MX]; // 0 == H 1 == P 2 == S

int main() {
	setIO("hps");

	int N, K;

	cin >> N >> K;

	for(int i = 0; i < N; i++) {
		char a; cin >> a;
		if(a == 'H') A[i] = 0;
		if(a == 'P') A[i] = 1;
		if(a == 'S') A[i] = 2;
	}

	// either she switches to h or p or s or stays

	for(int i = 0; i < N; i++) {
		for(int j = 0; j <= K; j++) {
			for(int k = 0; k < 3; k++) {
				if(k == A[i]) dp[i][j][k]++;
				dp[i+1][j+1][0] = max(dp[i+1][j+1][0], dp[i][j][k]); // switch to not item
				dp[i+1][j+1][1] = max(dp[i+1][j+1][1], dp[i][j][k]); // switch to not item
				dp[i+1][j+1][2] = max(dp[i+1][j+1][2], dp[i][j][k]); // switch to not item
				dp[i+1][j][k] = max(dp[i+1][j][k], dp[i][j][k]); // stay
			}
		}
	}

	int ret = 0;

	for(int i = 0; i < 3; i++) {
		ret = max(ret, dp[N-1][K][i]);
	}

	cout << ret << endl;

}

which I copied from this site.

Can anyone please explain to me what are we trying to do in the middle of the for loop?

[ I am new to dp, so an elaborate solution would be appreciated! ]

Thank you.

#2

Are you familiar with the concept of DP itself?

#3

Yes, I have gone through the DP module of the famous Competitive Programmer’s Handbook by Antti Laaksonen.

#4

Well, in the middle of the for loop we’re simply processing the new round, and checking the maximum number of games she can win given this new round.