HDU 3613 Best Reward (扩展KMP）

Best Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 238    Accepted Submission(s): 89

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 ≤ 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 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

 

Source

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

 

Recommend

lcy

 

 

题目大意：
给个字符串S，要把S分成两段T1,T2，每个字母都有一个对应的价值，如果T1，T2是回文串（从左往右或者从右往左读，都一样），那么他们就会有一个价值，这个价值是这个串的所有字母价值之和，如果不是回文串，那么这串价值就为0。问最多能获得多少价值？

 

对于我们只需要枚举扫描一遍extend数组，扫描到的当前位置之前为前半部分T1， 然后用根据extend数组可以判断T1是否是回文。那后半部分T2呢？  刚才是用S去匹配T， 如果要求后缀，只需要用T去匹配S，再得到一个数组extend2即可，根据这个extend2判断后半部分T2是否是回文。

 

一定要分成两部分噢，不能不分。如样例1

 

//============================================================================
// Name        : HDU3613.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
using namespace std;

void EKMP(char s[],char t[],int next[],int extend[])//s为主串，t为模板串
{
    int i,j,p,L;
    int lens=strlen(s);
    int lent=strlen(t);
    next[0]=lent;
    j=0;
    while(j+1<lent && t[j]==t[j+1])j++;
    next[1]=j;

    int a=1;
    for(i=2;i<lent;i++)
    {
        p=next[a]+a-1;
        L=next[i-a];
        if(i+L<p+1)next[i]=L;
        else
        {
            j=max(0,p-i+1);
            while(i+j<lent && t[i+j]==t[j])j++;
            next[i]=j;
            a=i;
        }
    }

    j=0;
    while(j<lens && j<lent && s[j]==t[j])j++;
    extend[0]=j;
    a=0;
    for(i=1;i<lens;i++)
    {
        p=extend[a]+a-1;
        L=next[i-a];
        if(L+i<p+1)extend[i]=L;
        else
        {
            j=max(0,p-i+1);
            while(i+j<lens && j<lent && s[i+j]==t[j])j++;
            extend[i]=j;
            a=i;
        }
    }
}

const int MAXN=500010;
char str1[MAXN],str2[MAXN];
int sum[MAXN];
int v[27];
int next[MAXN];
int extend1[MAXN],extend2[MAXN];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        for(int i=0;i<26;i++)
            scanf("%d",&v[i]);
        scanf("%s",str1);
        int len=strlen(str1);
        sum[0]=0;
        for(int i=0;i<len;i++)
        {
            sum[i+1]=sum[i]+v[str1[i]-'a'];
            str2[i]=str1[len-1-i];
        }
        str2[len]=0;
        EKMP(str2,str1,next,extend1);
        EKMP(str1,str2,next,extend2);
        int ans=-10000;
        //需要保证分成两部分，所以i从1到len-1
        for(int i=1;i<len;i++)
        {
            int tmp=0;
            if(i+extend1[i]==len)
            {
                tmp+=sum[len-i];
            }
            int pos=len-i;
            if(pos+extend2[pos]==len)
            {
                tmp+=sum[len]-sum[pos];
            }
            if(tmp>ans)ans=tmp;
        }
        printf("%d\n",ans);
    }
    return 0;
}