hdu 3613 Best Reward(manacher)

Best Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3215    Accepted Submission(s): 1324

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

  

   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
  

 

Source

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

题意：

一串项链是由26个不同的宝石组成的，每个宝石有不同的价值。如果项链是回文串的话，那么它的价值就是项链上所有宝石加起来的价值，否则价值为0。现在要将项链分成两半，然后算出它最大能值多少。

思路：

用manacher跑一边，如果以该点为回文中心能到达最左端，那么把该回文长度的位置记录下来，说明该回文能够到达最左端。然后如果能到达最右端，依然把从右边数到左的回文长度的位置记录下来，说明该回文能够达到最右端。然后枚举所有点，看是否能满足以该点i出发能到达最左端并且i+1的回文串能够到达最右端。并且记录一下项链的最大值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1000050;
int N,maxr[maxn],v[27],sum[maxn];
int pre[maxn],pos[maxn];
char a[maxn],b[maxn];
void init()
{
    int i;
    for(i=0;a[i];i++)
    {
        b[i*2+1]='#';
        b[i*2+2]=a[i];
    }
    N=2*i+1;
    b[0]='$',b[N]=b[N+1]='#';
}
void solve(int len)
{
    int id,maxn=0;
    for(int i=1;i<=N;i++)
    {
        maxr[i]=i<maxn?min(maxn-i,maxr[2*id-i]):1;
        while(b[i+maxr[i]]==b[i-maxr[i]]) maxr[i]++;
        if(i+maxr[i]>maxn) maxn=i+maxr[i],id=i;
        if(i-maxr[i]==0)
        {
            pre[maxr[i]-1]=1;
        }
        if(i+maxr[i]==2*len+2)
        {
            pos[maxr[i]-1]=1;
        }
    }
    int maxsum=-inf,tmp=0;
    for(int i=1;i<len;i++)
    {
        if(pre[i])
        {
            tmp+=sum[i];
        }
        if(pos[len-i])
        {
            tmp+=sum[len]-sum[i];
        }
        if(tmp>maxsum)
            maxsum=tmp;
        tmp=0;
    }
    printf("%d\n",maxsum);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(pre,0,sizeof(pre));
        memset(pos,0,sizeof(pos));
        for(int i=0;i<26;i++)
            scanf("%d",&v[i]);
        scanf("%s",a);
        int len=strlen(a);
        sum[0]=0;
        for(int i=1;i<=len;i++)
            sum[i]=sum[i-1]+v[a[i-1]-'a'];
        init();
        solve(len);
    }
    return 0;
}