codeforces-632B-Alice, Bob, Two Teams

632B-Alice, Bob, Two Teams

                time limit per test1.5 seconds memory limit per test256 megabytes

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.

input
5
1 2 3 4 5
ABABA
output
11

input
5
1 2 3 4 5
AAAAA
output
15

input
1
1
B
output
1

题目链接：cf-632B

题目大意：B能改变任意长度前缀或者后缀（即A->B,B->A）,问，如何改变使得B能获得的分数最大

题目思路：巧妙的暴力。

以下是代码：

#include<bits\stdc++.h>
#define ll long long
using namespace std;
ll a[500005];
ll b[500005];
ll k[500005];
int main(){
    int n;
    string s;
    cin >> n;
    for( int i=1 ; i<=n ; i++ ){
        scanf("%I64d", &k[i]);
    }
    cin >> s;
    for( int i=1 ; i<=n ; i++ ){
        if( s[i-1]=='A' ) a[i] = a[i-1]+k[i], b[i] = b[i-1];
        else b[i] = b[i-1]+k[i], a[i] = a[i-1];
    }
    ll ansmax = 0;
    for( int i=1 ; i<=n ; i++ ){
        ansmax = max( ansmax, a[i]+b[n]-b[i]);
        ansmax = max( ansmax, b[i]+a[n]-a[i]);
    }
    cout << ansmax << endl;
    return 0;
}