HDU 3616 Best Reward (Manacher算法 前綴回文＋後綴回文)

 

Best Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 785 Accepted Submission(s): 338

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


題目鏈接：www.Bkjia.com

題目答疑：輸入26個數字，分別表示26個字母的值，輸入一個字符串，將這個字符串分成兩半，若子串是回文串則其值為各字母之和否則為0，求所能得到的最大值

題目分析：先計算出字符串的前綴和，利用Manacher算法算出字符串的前綴回文序列和後綴回文序列並標記，然後每次加起來比較更新即可，實在感歎Manacher算法的強大！

#include 
#include 
#include 
using namespace std;
int const MAX = 500005;
char s[MAX << 1];
int p[MAX << 1], val[30], sum[MAX], len;
bool suf[MAX], pre[MAX];

void Manacher()
{
	int maxp = 0, maxl = 0;
	for(int i = len; i >= 0; i--)
	{
		s[i * 2 + 2] = s[i];
		s[i * 2 + 1] = '#';
	}
	s[0] = '*';
	for(int i = 2; i < 2 * len + 1; i++)
	{
		if(p[maxp] + maxp > i)
			p[i] = min(p[2 * maxp - i], p[maxp] + maxp - i);
		else 
			p[i] = 1;
		while(s[i - p[i]] == s[i + p[i]])
			p[i]++;
		if(p[maxp] + maxp < i + p[i])
			maxp = i;
		if(i - p[i] == 0)
			pre[p[i] - 1] = true;
		if(i + p[i] == 2 * len + 2)
			suf[p[i] - 1] = true;
	}
}

int main()
{
	int T;
	scanf(%d, &T);
	while(T--)
	{
		int ans = -0xfffffff;
		memset(suf, false, sizeof(suf));
		memset(pre, false, sizeof(pre));
		memset(sum, 0, sizeof(sum));
		for(int i = 0; i < 26; i++)
			scanf(%d, &val[i]);
		scanf(%s, s);
		len = strlen(s);
		for(int i = 1; i <= len; i++)
			sum[i] = sum[i - 1] + val[s[i - 1] - 'a'];
		Manacher();
		for(int i = 1; i < len; i++)
		{
			int tmp = 0;
			if(pre[i])
				tmp += sum[i];
			if(suf[len - i])
				tmp += sum[len] - sum[i];
			ans = tmp > ans ? tmp : ans;
		}
		printf(%d
, ans);
	}
}