HDU 3294 Girls' research (Manacher算法 + 记录区间)

Girls' research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 566    Accepted Submission(s): 212

Problem Description

One day, sailormoon girls are so delighted that they intend to research about palindromic strings. Operation contains two steps:
First step: girls will write a long string (only contains lower case) on the paper. For example, "abcde", but 'a' inside is not the real 'a', that means if we define the 'b' is the real 'a', then we can infer that 'c' is the real 'b', 'd' is the real 'c' ……, 'a' is the real 'z'. According to this, string "abcde" changes to "bcdef".
Second step: girls will find out the longest palindromic string in the given string, the length of palindromic string must be equal or more than 2.

 

Input

Input contains multiple cases.
Each case contains two parts, a character and a string, they are separated by one space, the character representing the real 'a' is and the length of the string will not exceed 200000.All input must be lowercase.
If the length of string is len, it is marked from 0 to len-1.

 

Output

Please execute the operation following the two steps.
If you find one, output the start position and end position of palindromic string in a line, next line output the real palindromic string, or output "No solution!".
If there are several answers available, please choose the string which first appears.

 

Sample Input

b babd a abcd

 

Sample Output

0 2 aza No solution!

 

Author

wangjing1111

 

Source

2010 “HDU-Sailormoon” Programming Contest

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3294

题目大意：输入一个字符ch和一个字符串，问如果把ch当作'a'的话，字符串的每个字符也要做相应变化，如b aa，若b为'a'，则b前面的a就为'a'前面的'z'，这里是循环表示，输出字符串的最长回文子串，如果最长回文子串串长为1，输出No solution!

题目分析：字符串根据要求变换一下，跑一次Manacher，每次更新maxl时，分别记录左右端点在原串中的位置

#include 
#include 
#include 
using namespace std;
int const MAX = 200005;
char s[MAX << 1], save[MAX << 1];
int p[MAX << 1], l, r;

int Manacher()
{
	int len = strlen(s), 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(maxl < p[i])
		{
			l = (i - p[i]) / 2;
			r = (i + p[i]) / 2 - 2;
			maxl = p[i];
		}
	}
	return maxl - 1;
}

int main()
{
	int ans;
	char ch[2];
	while(scanf("%s %s",ch, s) != EOF)
	{
		int len = strlen(s);
		l = r = 0;
		for(int i = 0; i < len; i++)
		{
			int tmp = s[i] - ch[0];
			if(tmp < 0)
				tmp = 26 + tmp;
			s[i] = 'a' + tmp;
		}
		strcpy(save, s);
		int ans = Manacher();
		if(ans == 1)
			printf("No solution!\n");
		else
		{
			printf("%d %d\n", l, r);
			for(int i = l; i <= r; i++)
				printf("%c", save[i]);
			printf("\n");
		}
	}
}