HDU 3294 Girls 'Research (Manacher Algorithm + record interval), hdumanacher
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.
InputInput 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.
OutputPlease 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 babda abcd
Sample Output
0 2azaNo solution!
Author wangjing1111 Source 2010 "HDU-Sailormoon" Programming Contest
Question link: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 3294
Enter a character ch and a string. If ch is treated as 'A', each character in the string must be changed accordingly, for example, B aa, if B is 'A', a in front of B is 'Z' in front of 'A'. Here it is a loop representation, and the longest return substring of the output string is, if the maximum length of the input string is 1, No solution is output!
Question analysis: the string is changed according to requirements. A Manacher is run, and the left and right endpoints are recorded in the original string each time maxl is updated.
#include <cstdio>#include <cstring>#include <algorithm>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");}}}
