题解 CF1409F 【Subsequences of Length Two】

dp[i][j][k]表示前i个字符修改j个包含k个s2[1]的最优个数，不修改s1[i + 1]时更新dp[i + 1][j][k + s1[i + 1] == s2[1]],
修改时若更新为第一个更新dp[i + 1][j + 1][k + 1],否则更新dp[i + 1][j + 1][k + s2[1] == s2[2]]

#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define se second
#define fi first
#define debug(x) cout<<#x<<" is "<<x<<endl
const int N = 200 + 5;

using namespace std;
typedef long long ll;
int dp[N][N][N];
typedef pair<int,int> pii;
int ans;
int n,t;
char s1[N],s2[N];
/*
int compute(string s,int pos,int change){
	int cnt = 0;
	if()s[pos] = s2[change];
	for(int i = 0;i < s.size();i++){
		if(s[i] != s2[0])
			continue;
		for(int j = i + 1;j <s.size();i++)
			if(s[j] == s2[1])
				cnt++;
	}
	return cnt;
}*/
#define dbg 0
const int INF = 0x3f3f3f3f;
int main(){
    
    
	#if dbg
	freopen("input.txt","r",stdin);
	#endif
	cin>>n>>t;
	cin>>(s1 + 1)>>(s2 + 1);
	for(int i = 0;i <N;i++)
		for(int j = 0;j < N;j++)
			for(int k = 0;k < N;k++)
				dp[i][j][k] = -INF;
	dp[0][0][0] = 0;
	for(int i = 0;i < n;i++){
    
    
		for(int j = 0;j <= min(i,t);j++){
    
    
			for(int k = 0;k <= n;k++){
    
    	
				if(dp[i][j][k] == -INF)continue;
				int e1 = (s1[i + 1] == s2[1]);
				int e2 = (s1[i + 1] == s2[2]);
				int e12 = (s2[1] == s2[2]);
				dp[i + 1][j][k + e1] = max(dp[i + 1][j][k + e1],dp[i][j][k] + (e2?k:0));
				if( j < t){
    
    
					dp[i + 1][j + 1][k + 1] = max(dp[i + 1][j + 1][k + 1],dp[i][j][k] + (e12?k:0));
					dp[i + 1][j + 1][k + e12] = max(dp[i + 1][j + 1][k + e12],dp[i][j][k] + k);
					ans = max(max(dp[i + 1][j + 1][k + 1],dp[i + 1][j + 1][k + e12]),ans);
				}
				ans = max(dp[i + 1][j][k + e1],ans);
			}	
		}
	}
	cout<<ans<<endl;
	return 0;
}