CodeForces - 1409F Subsequences of Length Two

传送门：CodeForces - 1409F

思路分析

$dp[i][j][k]$表示前$i$个字符，修改了$j$次，有$k$个$t[1]$的答案，往$i+1$转移的时候枚举变或不变就行了，记得特判$t[1]==t[2]$的情况

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 222;

char s[N],t[N];
int dp[N][N][N];

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n,m;
    cin>>n>>m;
    cin>>s+1;
    cin>>t+1;
    mem(dp,-1);
    dp[0][0][0] = 0;
    if(t[1] == t[2]) {
        int sum = 0;
        for(int i=1; i<=n; i++) {
            if(s[i] == t[1]) sum++;
            else {
                if(m) {
                    m--;
                    sum++;
                }
            }
        }
        cout<<sum*(sum-1)/2<<endl;
        return 0;
    }

    for(int i=0; i<n; i++) {
        for(int j=0; j<=m; j++) {
            for(int k=0; k<=i; k++) {
                if(dp[i][j][k]!=-1) {
                    int now = dp[i][j][k];
                    if(s[i+1] == t[1]) {
                        dp[i+1][j][k+1] = max(dp[i+1][j][k+1],now);
                        dp[i+1][j+1][k] = max(dp[i+1][j+1][k],now+k);
                    } else if(s[i+1] == t[2]) {
                        dp[i+1][j+1][k+1] = max(dp[i+1][j+1][k+1],now);
                        dp[i+1][j][k] = max(dp[i+1][j][k],now+k);
                    } else {
                        dp[i+1][j+1][k+1] = max(dp[i+1][j+1][k+1],now);
                        dp[i+1][j+1][k] = max(dp[i+1][j+1][k],now+k);
                        dp[i+1][j][k] = max(dp[i+1][j][k],now);
                    }
                }
            }
        }
    }
    
    int ans = 0;
    for(int j=0;j<=m;j++) for(int k=0;k<=n;k++) ans = max(ans,dp[n][j][k]);
    cout<<ans<<endl;




    return 0;
}