Alice, Bob, Two Teams

// Problem: B. Alice, Bob, Two Teams
// Contest: Codeforces - Educational Codeforces Round 9
// URL: https://codeforces.com/problemset/problem/632/B
// Memory Limit: 256 MB
// Time Limit: 1500 ms
// 2022-03-18 10:01:50
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
#define int long long
int n;
int p[500005], a[500005], b[500005];
string s;

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i ++)    
        cin>> p[i];
    cin >> s;
    int ans = 0;
    for (int i = 1; i <= n; i ++)
        if (s[i - 1] == 'B') {
            b[i] = p[i];
        }
        else {
            a[i] = p[i];
        }
        for (int i = 1;i <= n; i ++) {
            a[i] += a[i - 1];
            b[i] += b[i - 1];
        }
    int suf = 0, id = n + 1;
    for (int i=n; i >= 1; i --) {
        if (a[n] - a[i - 1] - b[n] + b[i - 1] > suf) {
            id = i;
            suf = a[n]-a[i - 1] - b[n] + b[i - 1];
        }
    }
    if (suf > 0) {
        ans = b[id] + a[n] - a[id - 1];
    }
    else {
        ans = b[n];
    }
    int pre = 0;
     id = -1;
    for (int i = 1;  i <= n; i ++) {
        if (a[i] - b[i] > pre) {
            pre = a[i]- b[i];c
            id = i;
        }
    }
    if (pre> 0) {
        ans = max (ans, b[n] - b[id] + a[id]);
    }
    cout << ans << endl;
}
signed main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}      

题解:通过尝试小的样例可以发现我们遍历两个方向取的最好的id。最后答案取得最优解