HDU 3613 Best Reward

最新推荐文章于 2020-09-05 12:09:37 发布

ChiLuManXi

最新推荐文章于 2020-09-05 12:09:37 发布

阅读量739

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/ChiLuManXi/article/details/50739089

本博客探讨了一个问题，即如何通过切割一条由不同宝石组成的项链来获得最大的价值。详细介绍了Manacher算法在判断回文串方面的应用，以及如何通过遍历找到最佳切割点以实现价值最大化。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Best Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1361 Accepted Submission(s): 551

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

题意是给定一个只含26个字母的权值，然后给定一个字符串，问如何将该字符串分成两块，且让这两块的字符串的权值和最大。一个字符串权值大小的规则是判断其是否是一个回文串，如果是权值就是其长度，如果不是，权值为0。

回文串可以考虑使用Manacher算法，对整个串进行回文串的判定，然后遍历，对于每个二分点，判断其两边是否是回文串并标记。最后再次遍历找到最大值。

代码如下：

/*************************************************************************
	> File Name: Best_Reward.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: 2016年02月24日 星期三 21时42分37秒
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;
const int MAXN = 500010;
int inde[30];
char str[MAXN];
char Ma[MAXN * 2];
int Mp[MAXN * 2];
int len;
long long value[MAXN];
bool flaga[MAXN];
bool flagb[MAXN];

void Manacher(){
    int l = 0;
    Ma[l ++] = '$';
    Ma[l ++] = '#';
    for(int i  = 0; i < len; i ++){
        Ma[l ++] = str[i];
        Ma[l ++] = '#';
    }
    Ma[l] = 0;
    int mx = 0, id = 0;
    for(int i = 0; i < l; i ++){
        Mp[i] = mx > i ? min(Mp[2 * id - i], mx - i) : 1;
        while(Ma[i + Mp[i]] == Ma[i - Mp[i]])
            Mp[i] ++;
        if(i + Mp[i] > mx){
            mx = i + Mp[i];
            id = i;
        }
        if(i - Mp[i] == 0)
            flaga[Mp[i] - 1] = true;
        if(i + Mp[i] == len + len + 2)
            flagb[Mp[i] - 1] = true;
    }
}

int T;

int main(void){
    cin >> T;
    while(T --){
        for(int i = 0; i < 26; i ++){
            cin >> inde[i];
        }
        memset(flaga, false, sizeof(flaga));
        memset(flagb, false, sizeof(flagb));
        cin >> str;
        len = strlen(str);
        for(int i = 1; i <= len; i ++){
            value[i] = value[i - 1] + inde[str[i - 1] - 'a'];
        }
        Manacher();
        long long ans = 0;
        long long temp = 0;
        for(int i = 1; i < len; i ++){
            if(flaga[i] == true)
                temp += value[i];
            if(flagb[len - i] == true)
                temp += value[len] - value[i];
            ans = ans > temp ? ans : temp;   
            temp = 0;
        }
        cout << ans << endl;

    }
}

查看原文：http://chilumanxi.org/2016/02/25/hdu-3613-best-reward/