【Codeforces】-632B-Alice, Bob, Two Teams(模拟,思维)
2016-08-22 21:21
357 查看
B. Alice, Bob, Two Teams
time limit per test
1.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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
output
input
output
input
output
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.
题意::题目是说从前或从后选一个子串(必须包括首或者尾,长度任意),把A变B,B变A,然后计算B的总价值,使之最大。
题解:先统计不变时B的价值,然后从前从后各扫一遍就行了
sum1记录从前到后变化A B,值的变化。sum2记录从后到前变化......
如:从前到后扫一遍,i=3,那么sum1就继续改变(i=1,i=2 .A变成B,B变成A已经记录),再更新ans
time limit per test
1.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
题意::题目是说从前或从后选一个子串(必须包括首或者尾,长度任意),把A变B,B变A,然后计算B的总价值,使之最大。
题解:先统计不变时B的价值,然后从前从后各扫一遍就行了
sum1记录从前到后变化A B,值的变化。sum2记录从后到前变化......
如:从前到后扫一遍,i=3,那么sum1就继续改变(i=1,i=2 .A变成B,B变成A已经记录),再更新ans
#include<cstdio> #include<cstring> #include<vector> #include<algorithm> using namespace std; #define CLR(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f int a[500010]; char s[500010]; int main() { int n; while(~scanf("%d",&n)) { __int64 ans=0,sum1,sum2; for(int i=1;i<=n;i++) scanf("%d",&a[i]); scanf("%s",s+1); for(int i=1;i<=n;i++) { if(s[i]=='B') ans+=a[i]; } sum1=ans; sum2=ans; for(int i=1;i<=n;i++) //从前到后一个for循环,两个for会超时.要包括首以及i之前的字母那么sum1就一直更新就好 { if(s[i]=='A') sum1+=a[i]; else sum1-=a[i]; ans=max(ans,sum1); } for(int i=n;i>=1;i--) { if(s[i]=='A') sum2+=a[i]; else sum2-=a[i]; ans=max(ans,sum2); } printf("%I64d\n",ans); } return 0; }
相关文章推荐
- 【CodeForces】632B - Alice, Bob, Two Teams(模拟)
- CodeForces 632B- Alice, Bob, Two Teams
- 【CodeForces 632B】 Alice, Bob, Two Teams(暴力)
- CodeForces-632B Alice, Bob, Two Teams 【dp】
- codeforces 632B B. Alice, Bob, Two Teams(暴力)
- Codeforces 632B:Alice, Bob, Two Teams(英文题。。。)
- CodeForces 632 B.Alice, Bob, Two Teams(水~)
- Alice, Bob, Two Teams codeforces 632B
- Alice, Bob, Two Teams CodeForces - 632B
- Alice, Bob, Two Teams —维护前缀和和后缀和
- codeforce-【B. Alice, Bob, Two Teams】
- codeforces-632B-Alice, Bob, Two Teams
- CodeForces 632B Alice, Bob, Two Teams
- CodeForces 632B Alice, Bob, Two Teams(暴力)
- Codeforces 272D Dima and Two Sequences【思维+模拟】
- CodeForces-632B-Alice, Bob, Two Teams
- Educational Codeforces Round 9 B. Alice, Bob, Two Teams 前缀和
- Codeforces 632B Alice, Bob, Two Teams 【水题】
- (Educational Codeforces Round 9 )Alice, Bob, Two Teams(前缀和)
- cf#ECR 9-B. Alice, Bob, Two Teams-DP