B. Alice, Bob, Two Teams

题目连接：

http://www.codeforces.com/contest/632/problem/B

Description

Alice and Bob are playing a game. The game involves splitting up game pieces into two teams. There are n pieces, and the i-th piece has a strength pi.

The way to split up game pieces is split into several steps:

First, Alice will split the pieces into two different groups A and B. This can be seen as writing the assignment of teams of a piece in an n character string, where each character is A or B.
Bob will then choose an arbitrary prefix or suffix of the string, and flip each character in that suffix (i.e. change A to B and B to A). He can do this step at most once.
Alice will get all the pieces marked A and Bob will get all the pieces marked B.
The strength of a player is then the sum of strengths of the pieces in the group.

Given Alice’s initial split into two teams, help Bob determine an optimal strategy. Return the maximum strength he can achieve.

Input

The first line contains integer n (1 ≤ n ≤ 5·105) — the number of game pieces.

The second line contains n integers pi (1 ≤ pi ≤ 109) — the strength of the i-th piece.

The third line contains n characters A or B — the assignment of teams after the first step (after Alice’s step).

Output

Print the only integer a — the maximum strength Bob can achieve.

Sample Input

5
1 2 3 4 5
ABABA

Sample Output

11

Hint

题意

有n个物品，每个物品有一个分值

然后每个物品有一个标牌，如果标牌上是A，那么就属于A，是B，那么就属于B

现在你可以选择一个前缀或者后缀，然后将上面的A改成B，将B改成A

然后问你最多B能够得到多少

题解：

暴力维护前缀和就好了

维护A的前缀和，和B的前缀和

翻转其实就是把A的变成B，B的变成A而已

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5+7;
int n;
long long p[maxn];
char s[maxn];
long long sum1[maxn];
long long sum2[maxn];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lld",&p[i]);
    scanf("%s",s+1);
    for(int i=1;i<=n;i++)
    {
        sum1[i]=sum1[i-1];
        sum2[i]=sum2[i-1];
        if(s[i]=='A')
            sum1[i]+=p[i];
        else
            sum2[i]+=p[i];
    }
    long long ans = max(sum2[n],sum1[n]);
    for(int i=1;i<=n;i++)
    {
        ans = max(ans,sum2[n]-sum2[i]+sum1[i]);
        ans = max(ans,sum2[i]+sum1[n]-sum1[i]);
    }
    cout<<ans<<endl;
}