hdu 3613 Best Reward

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原文链接：http://www.cnblogs.com/TheRoadToTheGold/p/7039321.html

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解决一道算法题目，通过Manacher算法寻找字符串中的最长回文子串，以此来最大化分割后的两个回文串的价值。

Best Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
http://acm.hdu.edu.cn/showproblem.php?pid=3613

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: v ₁, v ₂, ..., v ₂₆ (-100 ≤ v _i ≤ 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 v ₁, the value of 'b' is v ₂, ..., 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

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

Source

2010 ACM-ICPC Multi-University Training Contest（18）——Host by TJU

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题意：

给出26个数，每个数代表一个小写字母的分数

一个字符串的积分=字母的分数和

将原串分为两部分，如果某一部分是回文串，加上串的分数

问最大得分

manacher得出前缀回文和后缀回文

然后枚举划分位置

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 500011

using namespace std;

int T,val[27];
int sum[N],len,m,ans;
int id,pos,x,p[N*2];
bool pre[N],suf[N];
char s[N],ss[N*2];
void manacher()
{
    ss[m=0]='!';
    for(int i=1;i<=len;i++)
    {
        ss[++m]='#';
        ss[++m]=s[i];
    }
    ss[++m]='#';
    ss[m+1]='@';
    id=pos=0;
    for(int i=1;i<m;i++)
    {
        if(pos>i) x=min(p[2*id-i],pos-i);
        else  x=1;
        while(ss[i+x]==ss[i-x]) x++;
        p[i]=x;
        if(i+x>pos) { pos=i+x; id=i; }
        if(i-p[i]==0) pre[p[i]-1]=true;
        if(i+p[i]==m+1) suf[p[i]-1]=true;
    }
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        memset(pre,0,sizeof(pre));
        memset(suf,0,sizeof(suf));
        memset(sum,0,sizeof(sum));
        ans=-0x7fffffff;
        for(int i=0;i<26;i++) scanf("%d",&val[i]);
        scanf("%s",s+1);
        len=strlen(s+1);
        for(int i=1;i<=len;i++) sum[i]=sum[i-1]+val[s[i]-'a'];
        manacher();
        for(int i=1;i<len;i++)
             ans=max(ans,sum[i]*pre[i]+(sum[len]-sum[i])*suf[len-i]);
        printf("%d\n",ans);
    }
}