Ciel and Duel

#include <bits/stdc++.h>

using namespace std;

int n, m;
vector<int> atk;
vector<int> def;
int ciel[101];
int ans;

int main(){
    //freopen("cielandduel.in", "r", stdin);

    cin >> n >> m;
    for(int i=0; i<n; i++){
        string a;
        int b;
        cin >> a >> b;
        if(a == "ATK"){
            atk.push_back(b);
        }else if(a == "DEF"){
            def.push_back(b);
        }
    }
    for(int i=0; i<m; i++){
        cin >> ciel[i];
    }
    sort(atk.begin(), atk.end());
    sort(def.begin(), def.end());
    sort(ciel, ciel+m);
    int aidx = 0;
    int didx = 0;
    int val = 0;
    //For each card ciel has, first check Jiro's defense cards and then attack cards.
    //If the card isn't large enough to be used for defense cards, then use it for attack cards.
    //Since it is in sorted order if a card doesn't work there isn't a possibility it could work for any other of
    //Jiro's defense or attack cards. If it doesn't work for any of Jiro's cards, 'then just assume the card
    //can be used for direct attack used directly without any defense cards or attack cards in the way from
    //Jiro's side. Final answer will only be valid if all of Jiro's defense and attack cards have been used.
    for(int i=0; i<m; i++){
        if(didx<def.size() && ciel[i]>def[didx]){
            didx++;
            continue;
        }
        if(aidx<atk.size() && ciel[i]>=atk[aidx]){
            val+=ciel[i]-atk[aidx];
            aidx++;
            continue;
        }
        val+=ciel[i];
    }
    if(didx == def.size() && aidx == atk.size()){
        ans = max(ans, val);
    }
    aidx = 0;
    val = 0;
    //For the case where only Jiro's attack cards are hit with Ciel's attack cards, if one of ciel's cards
    //doesn't work on a attack card, no more values can be added because there are no more of Ciel's
    //attack cards that would work on a single one of Jiro's attack cards which is valid because of the
    //sorted order of Ciel's and Jiro's cards. Need to keep index of Jiro's attack cards because if the
    //cases are done we should just end because we cannot add any more of Ciel's card values to the
    //value of total points.
    for(int i=m-1; i>=0 && aidx<atk.size(); i--){
        //cout << ciel[i] << '\n';
        if(ciel[i]>atk[aidx]){
            val+=ciel[i]-atk[aidx];
            aidx++;
        }else{
            break;
        }
    }
    ans = max(ans, val);
    cout << ans << '\n';

    return 0;
}