Girls' research（已完善的Manacher算法模板：输出最长回文子串）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=3294

Girls' research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 863    Accepted Submission(s): 335

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 babda abcd

 

Sample Output

0 2azaNo solution!

 

Author

wangjing1111

 

Source

2010 “HDU-Sailormoon” Programming Contest

 

AC code：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<map>#define LL long long#define MAXN 1000010using namespace std;char s1[MAXN],s2[MAXN];//s1为原字符串，s2为转换后的字符串int p[MAXN];//p[i]表示以s2[i]为中心(包括s2[i])的回文半径,对应s1的回文长度 char ans[MAXN];//原字符串的最长回文串 int len; int l,r;//最长回文子串对应原串T中的位置：l = (i - p[i])/2; r = (i + p[i])/2 - 2;void pre(){int i,j,k;len=strlen(s1);s2[0]='$';s2[1]='#';for(i=0;i<len;i++){s2[i*2+2]=s1[i];s2[i*2+3]='#';}len=len*2+2;s2[len]=0;}int Manacher(){int i;int mx=0;int id;int ans=0;for(i=1;i<len;i++){if(mx>i)p[i]=min(p[2*id-i],mx-i);elsep[i]=1;for(;s2[i+p[i]]==s2[i-p[i]];p[i]++);if(ans<p[i]){ans=p[i];l = (i - p[i])/2;//最长回文子串对应原串s1中的左位置 r = (i + p[i])/2 - 2;//最长回文子串对应原串s1中的右位置}//ans=max(ans,p[i]);if(p[i]+i>mx){mx=p[i]+i;id=i;}}return ans-1;}int main(){ char ch,ch2;while(~scanf("%c %s",&ch,s1)){getchar();//注意要getchar();否则读取错误！！！pre();if(Manacher()<2){printf("No solution!\n");}else{int cnt=0; printf("%d %d\n",l,r);for(int i=l;i<=r;i++){if(s1[i]-ch>=0){ans[cnt++]='a'+s1[i]-ch;}else{ans[cnt++]='z'-(ch-s1[i])+1;}}ans[cnt]='\0';puts(ans);}}return 0;}