$$$dp1[i][j+1] + dp2[i-1][j-1] == LCS$$$

Only the different number of final $$$a$$$ string formed matters. So, there might be a possibility that you are counting the same character at an index multiple times because that character may appear at different positions in string $$$b$$$.

You can create a $$$vis[i][j]$$$ array to mark $$$1$$$ if you have placed $$$jth$$$ character at index $$$i$$$, and then count the final answer using the number of elements that are marked $$$1$$$ in the $$$vis$$$ array.

#include<bits/stdc++.h>
#define ll int
#define f(i,a,b) for(ll i=a;i<b;i++)
#define mod 1000000007
#define mp make_pair
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define ff first
#define ss second
#define rf(i,a,b) for(ll i=a;i>=b;i--)
#define sc(a) scanf("%lld",&a)
#define pf printf
#define sz(a) (int)(a.size())
#define psf push_front
#define ppf pop_front
#define ppb pop_back
#define pb push_back
#define pq priority_queue
#define all(s) s.begin(),s.end()
#define sp(a) setprecision(a)
#define rz resize
#define ld long double
#define inf (ll)1e18
#define ub upper_bound
#define lb lower_bound
#define bs binary_search
#define eb emplace_back
const double pi = acos(-1);
ll binpow(ll a, ll b){ll res=1;while(b!=0){if(b&1)res*=a;a*=a;b>>=1;}return res;}
ll binpow(ll a, ll b, ll md){ll res=1;a%=md;if(a==0)return 0;while(b!=0){if(b&1)res*=a,res%=md;a*=a,a%=md;b>>=1;}return res%md;}
 
using namespace std;

ll dp[5003][5003],rdp[5003][5003],n,m;
string a,b;

ll fn(ll i, ll j)
{
    if(i>=n || j>=m)
        return 0;
    ll &ans=dp[i][j];
    if(ans==-1)
    {
        ans=0;
        if(a[i]==b[j])
            ans=1+fn(i+1,j+1);
        else
            ans=max({ans,fn(i+1,j),fn(i,j+1)});
    }
    return ans;
}

ll rfn(ll i, ll j)
{
    if(i<0 || j<0)
        return 0;
    ll &ans=rdp[i][j];
    if(ans==-1)
    {
        ans=0;
        if(a[i]==b[j])
            ans=1+rfn(i-1,j-1);
        else
            ans=max({ans,rfn(i-1,j),rfn(i,j-1)});
    }
    return ans;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int z=1;
    f(i,1,z+1)
    {
        cin>>a>>b;
        n=sz(a),m=sz(b);
        memset(dp,-1,sizeof(dp)),memset(rdp,-1,sizeof(rdp));
        ll ans=0,lcs=fn(0,0);
        vector<vector<bool> > vis(n+69,vector<bool> (269));
        f(i,0,n+1)
        {
            f(j,0,m)
            {
                if(rfn(i-1,j-1)+fn(i,j+1)==lcs)
                    vis[i][(int)b[j]]=1;
            }
        }
        f(i,0,n+1)
        {
            f(j,0,269)
            {
                if(vis[i][j])
                    ans++;
            }
        }
        cout<<ans<<"\n";
    }
}