HDU----（3294）Girls' research（manacher）

Girls' research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 537    Accepted Submission(s): 199

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

代码：题目意思:

给定一个字符，以该字符作为'a'字符，举列子： c abac , c=a; b=z ,a=y;

 然后找出他的最长回文子串，标出他的起始位置和最终位置...然后输出它的回文串（变换后的）

做法：  先用manacher求出它的最长回文串，算法起始点，可以考虑另外开一个数组，来存储原始字串....然后依据起始和结尾点来输出即可

，采用manacher算法处理回文子串

代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 #define maxn 400050
 5 char str[maxn];
 6 int rad[maxn];
 7 int Min(int a,int b){
 8   return a<b?a:b;
 9 }
10 void init(int len,char s[]){
11     memset(rad,0,sizeof(int)*(2*len+2));
12     s[len*2+2]='\0';
13     int i,j=1;
14   for(i=len*2+1;i>0;i--){
15       if(i&1) s[i]='#';
16   else{ s[i]=s[len-j];
17      j++;
18    }
19   }
20   s[0]='$'; //防止溢出
21 }
22 int manacher(int len){
23     int id,i,ans=0;
24     for(i=1;i<len*2+1;i++){
25      if(id+rad[id]>i)  rad[i]=Min(rad[id*2-i],id+rad[id]-i);
26      while(str[i-rad[i]]==str[i+rad[i]]) rad[i]++;
27      if(i+rad[i]>id+rad[id]) id=i;
28      if(ans<rad[i])ans=rad[i];
29     }
30     return ans;
31 }
32 int main(){
33   char sav[2];
34    int len,i;
35    //system("call test.in");
36    //freopen("test.in","r",stdin);
37    // fclose(stdin);
38    while(scanf("%s%s",sav,str)!=EOF){
39      len=strlen(str);
40      init(len,str);
41      int ans=manacher(len);
42      if(ans<=2)puts("No solution!");
43      else{
44      for(i=1;i<len*2+1;i++)
45          if(ans==rad[i]) break;
46          int st,en;
47          st=(i-ans)/2;
48          en=st+ans-2;
49         printf("%d %d\n",st,en);
50       for(int j=(st+1)*2;j<(st+ans)*2;j+=2){
51             if(str[j]-(sav[0]-'a')<'a')
52                printf("%c",str[j]+('z'-sav[0]+1));
53             else
54                printf("%c",str[j]-(sav[0]-'a'));
55       }
56       puts("");
57      }
58    }
59   return 0;
60 }