HDU 3613

最新推荐文章于 2020-09-05 12:09:37 发布

原创最新推荐文章于 2020-09-05 12:09:37 发布 · 196 阅读

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#manacher

本文探讨了一种算法，用于解决将军的项链问题。通过Manacher算法和枚举策略，该算法能有效找出将项链分割成两个部分的最佳方式，使得这两部分作为回文串时的价值总和最大。输入包括宝石种类的价值和代表项链的字符串，输出为最大价值。

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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

这道题题意就是先给你26个数字，代表的是a~z这26种宝石的价格，然后给你一个字符串，代表的是这串宝石，让你把他分成两段，分开后如果这段宝石是一个回文串，那么他的价值就是每一个宝石的价值之和，如果不是回文串，他的价值就是0，这道题的话用manacher+枚举就可以a了

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<set>
#define mod 10007
using namespace std;
int v[30];
char s[500005],s1[1000005];
int p[1000005],sum[1000005];
int manacher()
{
    s1[0]='$';
    s1[1]='#';
    int k=2;
    int n=strlen(s);
    for(int a=0;a<n;a++)
    {
        s1[k++]=s[a];
        s1[k++]='#';
    }
    s1[k]='\0';
    int id=0,mx=0;
    for(int i=1;i<k;i++)
    {
        if(i<mx)
        {
            p[i]=min(mx-i,p[2*id-i]);
        }
        else
          p[i]=1;
        while(s1[i-p[i]]==s1[i+p[i]])
            p[i]++;
        if(i+p[i]>mx)
        {
            id=i;
            mx=i+p[i];
        }

    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int a=0;a<26;a++)
        {
            scanf("%d",&v[a]);
        }
        scanf("%s",s);
        manacher();
        int n=strlen(s),ans=0;
        sum[0]=v[s[0]-'a'];
        for(int a=1;a<n;a++)
        {
            sum[a]=sum[a-1]+v[s[a]-'a'];
//            printf("%d ",sum[a]);
        }
//        printf("\n");
        for(int a=0;a<n-1;a++)
        {
            int num=0;
            int k1=a+1,k2=n-a-1;   //k1表示的是前段长度，k2是后段的长度
            int k=p[a+2]-1;
            if(k1==k)
            {
                num+=sum[a];
            }
            k=p[a+n+2]-1;
            if(k2==k)
            {

                num+=(sum[n-1]-sum[a]);
            }
            ans=max(ans,num);
        }
        printf("%d\n",ans);
    }
    return 0;
}