HDU 3613 Best Reward

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

 

题意：有一串项链是由26中宝石构成的（26个英文字母），每种宝石都有它自身的价值（可正可负），现在知道这条项链是由什么宝石构成的了，需要计算它的价值最大（计算的要求是把这个串分成两个串，如果这两个串都是回文串，那么最大价值就是宝石价值总和，如果不是，那就计算回文串最大的价值总和，这个回文串必须接首位或者尾位）。

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

const int N=1e6+10;
const int INF=0x3f3f3f3f;

char S[N], s[N];
int r[N], a[30], sum[N], L[N], R[N], k; ///R数组标记接首位的回文串（下标是回文串长度），L数组标记接尾位的回文串

void Manacher()
{
    int i, mx, id = 0;

    for (i = 2; i < k; i++)
    {
        r[i] = 1;

        mx = r[id]+id;

        if (mx > i) r[i] = min(r[2*id-i], mx-i);

        while (s[i-r[i]] == s[i+r[i]]) r[i]++;

        if (mx < r[i]+i) id = i;

        if (r[i] == i) L[r[i]-1] = 1; ///因为r保存的是半径，如果半径等于i，那么肯定接首位
        if (r[i]+i == k) R[r[i]-1] = 1; ///r[i]+i是以i为中心的回文串的边界，如果等于k，则接尾位
    }
}

int main()
{
    int T, i, ls, ans, num;

    scanf("%d", &T);

    while (T--)
    {
        k = 2;

        memset(r, 0, sizeof(r));
        memset(sum, 0, sizeof(sum));
        memset(L, 0, sizeof(L));
        memset(R, 0, sizeof(R));

        for (i = 0; i < 26; i++)
            scanf("%d", &a[i]);
        scanf("%s", S);

        ls = strlen(S);

        for (i = 1; i <= ls; i++)
            sum[i] = sum[i-1] + a[S[i-1]-'a']; ///先将价值之和保存下来

        s[0] = '$'; s[1] = '#';
        for (i = 0; S[i] != '\0'; i++)
        {
            s[k++] = S[i];
            s[k++] = '#';
        }

        Manacher();

        ans = a[S[0]-'a'];

        for (i = 1; i < ls; i++)
        {
            num = 0;

            if (L[i]) num = sum[i];
            if (R[ls-i]) num += (sum[ls]-sum[i]);

            ans = max(ans, num);
        }

        printf("%d\n", ans);
    }

    return 0;
}

转载于:https://www.cnblogs.com/syhandll/p/4871177.html