HDU 3294 Girls ' Manaler algorithm

Surface:

One day, Sailormoon Girls is so delighted the they intend to the, about palindromic strings. Operation contains, steps:
First Step:girls would write a long string (only contains lower case) on the paper. For example, "ABCDE", but ' a ' inside was not the real ' a ', so means if we define the ' B ' is the real ' a ' and then we can inf Er that ' C ' are the real ' B ', ' d ' is the real ' C ' ..., ' a ' is the real ' z '. According to this, the string "ABCDE" Changes to "Bcdef".
Second Step:girls'll 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 parts, a character and a string, they is separated by one space, the character representing the RE Al ' A ' is and the length of the string won't exceed 200000.All input must be lowercase.
If the length of string is Len, it's marked from 0 to len-1.
Output
Please execute the operation following the steps.
If you find one, output the start position and end position of palindromic string in a line, next line output the real pal Indromic string, or output "No solution!".
If There is several answers available, please choose the string which first appears.
Sample Input
b babd
a ABCD
Sample Output
0 2
Aza
No solution!

Main topic:

Now there is a string with some palindrome strings inside. Now let you find out which one is the longest and output.
But there is another condition, each string preceded by a character that represents a set of substitution rules.
If it's B, then it's \ (b->a,c->b,d->c...a->z\)
Finally, the replacement string is output and the left and right bounds of the string are output (character numbering starts at 0).

General idea:

If there is no replacement rule, it is a manaler bare template. Then, the longest palindrome string to be obtained and the relative position of some operations.
This operation is divided into when \ (pos\) is an odd or even number.
Vo specific operation look at the code

//pos是定位的最长回文串的位置，len是对应的长度if(pos%2){    l=pos-(len-1);    l/=2;    --l;    r=pos+(len-1);    r/=2;    --r;}else{    mid=pos/2;    l=mid-len/2-1;    r=mid+len/2-1;}

Code:

#include <bits/stdc++.h>using namespace Std;const int Maxn=2e5+7;char str[maxn],tran[maxn<<2],basc;int p[    maxn<<2];void print (int len) {for (int i=1;i<len;++i) cout<<p[i]<< ""; Cout<<endl;}    void Manaler () {memset (tran,0,sizeof (Tran));    memset (P,0,sizeof (p));    int Len_1=strlen (str), len_2=1;    Tran[0]= ' $ ';        for (int i=0;i<len_1;++i) {tran[len_2++]= ' # ';    Tran[len_2++]=str[i];    } tran[len_2++]= ' # ';    tran[len_2]= ' + ';    int mi=0,r=0;        for (int i=1;i<len_2;++i) {if (i<=r) p[i] = min (r-i,p[2*mi-i]);        else p[i]=1;        while (Tran[i-p[i]]==tran[i+p[i]]) ++p[i];            if (i+p[i]>r) {r = I+p[i];        Mi=i;    }}//print (len_2);        int k=basc-' a ', pos=0,len=0;                for (int i=1;i<len_2;++i) {if (P[i]-1>len) {len=p[i]-1;            Pos=i; }} if (LEN&LT;2) cout<< "No solution!"        <<endl;            else{int mid,l,r;                if (pos%2) {l=pos-(len-1);                l/=2;                --l;                r=pos+ (len-1);                r/=2;            --r;                }else{MID=POS/2;                l=mid-len/2-1;            r=mid+len/2-1;            } cout<<l<< "" <<r<<endl;                    for (int i=pos-len;i<=pos+len;++i) {if (tran[i]!= ' # ') {char c= (tran[i]-k);                    if (c< ' a ') c+=26;                cout<<c;        }} cout<<endl;    }}int Main () {Ios::sync_with_stdio (false);    while (CIN&GT;&GT;BASC&GT;&GT;STR) {Manaler (); } return 0;}

HDU 3294 Girls ' Manaler algorithm