标签：manacher

Best Reward

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

1
6

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这题我是用manacher做的,不过很多人好像是EKMP搞的

先求出p数组,然后遍历一遍,找到回文到字符串端点的回文串,然后求最大值即可

/*************************************************************************
    > File Name: hdu3613.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月02日 星期一 11时59分27秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 1100000;
char str[N];
char mat[N];
int p[N];
int val[30];
int sum[N];

void manacher (int cnt)
{
	memset (p, 0, sizeof(p));
	int MaxId = 0, id;
	for (int i = 1; i < cnt; ++i)
	{
		if (MaxId > i)
		{
			p[i] = min (p[2 * id - i], MaxId - i);
		}
		else
		{
			p[i] = 1;
		}
		while (str[i + p[i]] == str[i - p[i]])
		{
			++p[i];
		}
		if (p[i] + i > MaxId)
		{
			MaxId = p[i] + i;
			id = i;
		}
	}
}

int main ()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		for (int i = 0; i < 26; ++i)
		{
			scanf("%d", &val[i]);
		}
		scanf("%s", mat);
		int cnt = 2;
		int len = strlen (mat);
		for (int i = 0; i < len; ++i)
		{
			str[cnt++] = mat[i];
			str[cnt++] = '#';
		}
		str[0] = '?';
		str[1] = '#';
		str[cnt] = '\0';
		sum[0] = 0;
		manacher (cnt);
		for (int i = 1; i < cnt; ++i)
		{
			sum[i] = sum[i - 1];
			if (str[i] != '#')
			{
				sum[i] += val[str[i] - 'a'];
			}
		}
		int ans = 0;
		for (int i = 2; i < cnt - 1; ++i)
		{
			if (p[i] - 1 == len)
			{
				continue;
			}
			if (i - p[i] + 1 == 1)
			{
				int tmp = sum[i + p[i] - 1];
				int id = (cnt - 1 + i + p[i] - 1) / 2;
				if (p[id] == cnt - 1 - id + 1)
				{
					tmp += sum[cnt - 1] - sum[i + p[i] - 2];
				}
				ans = max(ans, tmp);
			}
			else if (i + p[i] - 1 == cnt - 1)
			{
				int tmp = sum[i + p[i] - 1] - sum[i - p[i]];
				int id = (1 + i - p[i] + 1) / 2;
				if (p[id] == id)
				{
					tmp += sum[i - p[i] + 1];
				}
				ans = max(ans, tmp);
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}

hdu3613---Best Reward

标签：manacher

原文地址：http://blog.csdn.net/guard_mine/article/details/43406937