2014
11-27

# Best Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 364    Accepted Submission(s): 144

Problem Description
After an uphill battle, General Li won a great victory. Now the head of state decide to reward him with honor and treasures for his great exploit.

One of these treasures is a necklace made up of 26 different kinds of gemstones, and the length of the necklace is n. (That is to say: n gemstones are stringed together to constitute this necklace, and each of these gemstones belongs to only one of the 26 kinds.)

In accordance with the classical view, a necklace is valuable if and only if it is a palindrome – the necklace looks the same in either direction. However, the necklace we mentioned above may not a palindrome at the beginning. So the head of state decide to
cut the necklace into two part, and then give both of them to General Li.

All gemstones of the same kind has the same value (may be positive or negative because of their quality – some kinds are beautiful while some others may looks just like normal stones). A necklace that is palindrom has value equal to the sum of its gemstones’
value. while a necklace that is not palindrom has value zero.

Now the problem is: how to cut the given necklace so that the sum of the two necklaces’s value is greatest. Output this value.

Input
The first line of input is a single integer T (1 ≤ T ≤ 10) – the number of test cases. The description of these test cases follows.

For each test case, the first line is 26 integers: v1, v2, …, v26 (-100 ≤ vi ≤ 100, 1 ≤ i ≤ 26), represent the value of gemstones of each kind.

The second line of each test case is a string made up of charactor ‘a’ to ‘z’. representing the necklace. Different charactor representing different kinds of gemstones, and the value of ‘a’ is v1, the value of ‘b’ is v2, …, and so on.
The length of the string is no more than 500000.

Output
Output a single Integer: the maximum value General Li can get from the necklace.

Sample Input
2
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
aba
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
acacac

Sample Output
1
6


#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;

const int MAX=500000+10;
char s1[MAX],s2[MAX];
int next[MAX],extend1[MAX],extend2[MAX];
int sum[MAX],val[27];

void get_next(char *a,int len){
int k=0,i=1;
next[0]=len;//可有可无,因为用不上
while(k+1<len && a[k] == a[k+1])++k;
next[1]=k;//这里预先算好next[1]是因为不能k=0,否则next[i-k]=next[i]不是已算好的
k=1;
while(++i<len){//和EKMP的过程一样
int maxr=k+next[k]-1;
next[i]=min(next[i-k],max(maxr-i+1,0));//这里是扩展KMP的精髓,即算法核心思想就是这
while(i+next[i]<len && a[next[i]] == a[i+next[i]])++next[i];
if(i+next[i]>k+next[k])k=i;
}
}

void EKMP(char *a,char *b,int *extend,int len){
get_next(a,len);
int k=0,i=0;
while(k<len && a[k] == b[k])++k;
extend[0]=k;
k=0;
while(++i<len){
int maxr=k+extend[k]-1;
extend[i]=min(next[i-k],max(maxr-i+1,0));//next[i-k]是a与b从i开始的可能已经匹配的长度
while(i+extend[i]<len && a[extend[i]] == b[i+extend[i]])++extend[i];//这里是扩展KMP的精髓,即算法核心思想就是这
if(i+extend[i]>k+extend[k])k=i;
}
}

int main(){
int n;
cin>>n;
while(n--){
for(int i=0;i<26;++i)cin>>val[i];
scanf("%s",s1);
int len=strlen(s1);
for(int i=1;i<=len;++i){
sum[i]=sum[i-1]+val[s1[i-1]-'a'];
s2[i-1]=s1[len-i];
}
EKMP(s1,s2,extend1,len);
EKMP(s2,s1,extend2,len);
int ans=0,temp=0;
for(int i=1;i<len;++i){
if(extend1[len-i] == i)temp+=sum[i];//表示前i个字符是回文串
if(extend2[i] == len-i)temp+=sum[len]-sum[i];//表示后len-i个字符为回文串
if(temp>ans)ans=temp;
temp=0;
}
cout<<ans<<endl;
}
return 0;
}