【CodeForces】 632B Alice, Bob, Two Teams

游戏分配策略优化

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本文介绍了一个涉及游戏元素分配的问题，玩家Alice将游戏元素分为两组，Bob通过翻转某一前缀或后缀来最大化自己的得分。文章提供了一个解决此问题的算法实现。

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.

Examples

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

Note

In the first sample Bob should flip the suffix of length one.

In the second sample Bob should flip the prefix or the suffix (here it is the same) of length 5.

In the third sample Bob should do nothing.

这道题让我明白了什么叫英语不好举步维艰，大概题意是bob只能改变字母的前缀（从后面开始改变字母由A->B或由B->A，个数不限）或后缀（同上），bob得到所有B字符对应的数字的和，问bob能得到的数字和最大是多少。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
const int N = 5e5+10;
LL a[N];
char c[N];
int main() {
	int n;
	while(scanf("%d",&n)!=EOF) {
		memset(a,0,sizeof(a));
		for(int i=0; i<n; i++) {
			scanf("%lld",&a[i]);
		}
		scanf("%s",c);
		LL sum1=0,sum2=0;
		LL maxn1=0,maxn2=0;
		LL temp=0,ans;
		for(int i=0; i<n; i++) {
			if(c[i]=='B')
				temp+=a[i];
		}
		//	printf("%lld==\n",temp);
		for(int i=0; i<=n; i++) {               
			if(i==0) {
				sum1=temp;
			} else if(c[i-1]=='A') {      //此处细节处理不当wa了一次i-1；
				sum1+=a[i-1];
			} else if(c[i-1]=='B') {
				sum1-=a[i-1];
			}
			maxn1=max(maxn1,sum1);
		}
		for(int i=n; i>=0; i--) {
			if(i==n) {
				sum2=temp;
			} else if(c[i]=='A') {
				sum2+=a[i];
			} else if(c[i]=='B') {
				sum2-=a[i];
			}
			maxn2=max(maxn2,sum2);
		}
		//	printf("%lld %lld---\n",maxn1,maxn2);
		ans=max(maxn1,maxn2);
		printf("%lld\n",ans);
	}
	return 0;
}