hdu 3294 Girls' research Manacher回文串

Girls' research(Manacher)

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3294

Appoint description: System Crawler  (2016-01-17)

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!                

 

题意：给出一个加密的字符串，然后找字符串中的最长回文字串的起始位置和值

思路：先解密，再使用Manacher算法找出最长回文字串，其中有一个公式，新串中的位置( i , j )对应原始串的( (i-mp[i])/2 , (i+mp[i]-4)/2 )，很容易推导出来。

代码：

#include <iostream>#include <stdio.h>#include <string.h>#define MAX 200010using namespace std;char ma[MAX<<1];int mp[MAX<<1];int manacher(char s[],int len){    int L=0;    ma[L++]='$';    ma[L++]='#';    for(int i=0;i<len;i++)    {        ma[L++]=s[i];        ma[L++]='#';    }    ma[L]='\0';    int mx=0,id=0;    for(int i=0;i<L;i++)    {        mp[i]=mx>i?min(mp[2*id-i],mx-i):1;        while(ma[i+mp[i]]==ma[i-mp[i]]) mp[i]++;        if(i+mp[i]>mx)        {            mx=i+mp[i];            id=i;        }    }    return L;}int main(){    char s[MAX],ch[2];    int len,nlen,i,pre,maxp,maxn;    int bg,ed;    while(~scanf("%s%s",ch,s))    {        maxp=maxn=-1;        len=strlen(s);        pre=ch[0]-'a';        for(i=0;i<len;i++)        {            s[i]=s[i]-pre;            if(s[i]<'a')                s[i]+=26;        }        nlen=manacher(s,len);        for(i=1;i<nlen;i++)        {            if(maxn<mp[i])            {                maxp=i;                maxn=mp[i];            }        }        if(maxn-1<2)            printf("No solution!\n");        else        {            bg=(maxp-maxn)/2;//起始位置            ed=(maxp+maxn-4)/2;//结束位置            printf("%d %d\n",bg,ed);            for(i=bg;i<=ed;i++)                printf("%c",s[i]);            printf("\n");        }    }    return 0;}