hdu-3613 Best Reward (manacher算法)

Best Reward

题目链接

 

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

 

先manacher算法扫一遍，前缀和也扫一遍，暴力枚举前面的回文串中点，如果这个回文串的第一位的id不是1，那么这个回文串的值为0，不然就用前缀和算出前串总值，这样同理也可以得到后串的值，加起来以后就是把串分为两部分得到的值，然后和原来的答案比较。

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int maxn = 500100;
char s[maxn];
char txt[maxn<<1];
int p[maxn<<1];
int sum[maxn<<1];
int len; int val = 0; int n;
int v[26];
int oo = 0x3f3f3f3f;

void make() {
    txt[0] = '$';
    int pos = 0;
    for(int i = 0; i < len; i++) {
        txt[++pos] = '#';
        txt[++pos] = s[i];
    }txt[++pos] = '#';
    n = len * 2 + 2;
    txt[n] = '';
}

int get(int id) {
    if(txt[id] >= 'a' && txt[id] <= 'z') {
        return v[txt[id] - 'a'];
    }
    return 0;
}


void solve() {
    int maxx = 0, id;
    for(int i = 1; i < n; i++) {
        if(i < maxx)
            p[i] = min(p[id * 2 - i], p[id] + id - i);
        else p[i] = 1;
        while(txt[i - p[i]] == txt[i + p[i]]) {
            if(i + p[i] > maxx) {
                maxx = i + p[i];
                id = i;
            }
            p[i]++;
        }
    }
    for(int i = 1; i < n; i++) sum[i] = sum[i - 1] + get(i);
    for(int i = 1; i < n; i++) p[i]--;
    int ans = -oo, tsum = 0;
    for(int i = 2; i < n - 1; i++) {
        tsum = 0;
        int l, r;
        int newl, newr;

        if(p[i]) {
            l = i - p[i];
            r = i + p[i];
        }
        else {
            l = i - 1, r = i + 1;
        }

        if(l != 1 && r != len * 2 + 1) {
            tsum = 0;
            l = 1;
            newl = r, newr = len * 2 + 1;
        }
        else if(l != 1 && r == len * 2 + 1) {
            newl = 1, newr = l;
            tsum = sum[r] - sum[l];
        }
        else if(l == 1 && r != len * 2 + 1) {
            newl = r, newr = len * 2 + 1;
            tsum = sum[r] - sum[l];
        }
        else {
            tsum = -oo;
        }

        int newmid = (newl + newr) / 2;
        //printf("check %d %d %d %d %d %d %d
", l, r, newl, newr, tsum, ans, newmid);
        if(newmid - p[newmid] != 1 && newmid + p[newmid] != len * 2 + 1) {
            ans = max(ans, tsum);
            continue;
        }
        tsum += (sum[newr] - sum[newl]);
        ans = max(ans, tsum);
    }
    printf("%d
", ans);
}

int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        for(int i = 0; i < 26; i++) scanf("%d", &v[i]);
        scanf("%s", s);
        memset(p, 0, sizeof(p));
        memset(sum, 0, sizeof(sum));
        len = strlen(s);
        make();
        solve();
    }    
}