The Ninth Hunan Collegiate Programming Contest (2013) Problem A

Problem A

Almost Palindrome

Given a line of text, find the longest almost-palindrome substring. A string S is almost-palindrome if

1. S begins and ends with a letter, and
2. a(S) and b(S) have at most 2k positions with different characters

Here a(S) is the string after removing all non-letter characters and converting all the letters to lowercase, b(S) is the reversed string of a(S).

For example, when k=1, "Race cat" is almost-palindrome, because a(S)="racecat" and b(S)="tacecar" differ at exactly 2 positions.

Input

There will be at most 25 test cases. Each test case contains two lines. The first line is k (0<=k<=200). The second line contains a string with at least one letter and at most 1,000 characters (excluding the newline character). The string will only contain letters, spaces and other printable characters like ("," or "." etc) and will not start with a whitespace.

Output

For each test case, print the length of the longest almost-palindrome substring and its starting position (starting from 1). If there is more than one such string, print the smallest starting position.

Sample Input

1
Wow, it is a Race cat!
0
abcdefg
0
Kitty: Madam, I'm adam.

Output for the Sample Input

Case 1: 8 3
Case 2: 1 1
Case 3: 15 8

The Ninth Hunan Collegiate Programming Contest (2013)
Problemsetter: Rujia Liu
Special Thanks: Feng Chen, Md. Mahbubul Hasan

   这个题应该是由最长回文串改编而来的，最长回文串是K=0的情形，注意一个小地方，当diff=K的时候还是可以扩展的。程序应该清晰易懂，而不是来展示小聪明的。

#include <iostream>
#include <stdio.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#include <map>
#include <stack>
#include <math.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
const int Max_N=1008 ;
char old_str[Max_N] ;
char str[Max_N] ;
int K ;
int my_hash[Max_N] ;
int Len ;
struct Node{
    int Left_id ;
    int Lenght ;
    Node(){} ;
    Node(int id ,int len):Left_id(id),Lenght(len){} ;
};
Node gao_case1(int id){
     int Left ,Right ;
     Left=id-1 ;
     Right=id+1 ;
     int dif=0 ;
     int ans_left_id=my_hash[id] ,ans_length=1 ;
     while(Left>=0&&Right<Len&&dif<K){
         if(str[Left]!=str[Right])
               dif++ ;
         ans_left_id=my_hash[Left] ;
         ans_length=my_hash[Right]-my_hash[Left]+1 ;
         Left-- ;
         Right++ ;
     }
     /*k th will go on*/
     while(Left>=0&&Right<Len&&str[Left]==str[Right]){
         ans_left_id=my_hash[Left] ;
         ans_length=my_hash[Right]-my_hash[Left]+1 ;
         Left-- ;
         Right++ ;
     }
     return Node(ans_left_id,ans_length) ;
}

Node gao_case2(int id){
     int Left ,Right ;
     Left=id  ;
     Right=id+1  ;
     int dif=0 ;
     int ans_left_id=my_hash[id] ,ans_length=0 ;
     while(Left>=0&&Right<Len&&dif<K){
         if(str[Left]!=str[Right])
               dif++ ;
         ans_left_id=my_hash[Left] ;
         ans_length=my_hash[Right]-my_hash[Left]+1 ;
         Left-- ;
         Right++ ;
     }
     /*k th will go on*/
     while(Left>=0&&Right<Len&&str[Left]==str[Right]){
         ans_left_id=my_hash[Left] ;
         ans_length=my_hash[Right]-my_hash[Left]+1 ;
         Left-- ;
         Right++ ;
     }
     return Node(ans_left_id,ans_length) ;
}

int main(){
   int L_old ,ans_len ,ans_left_id ,T=1 ;
   while(scanf("%d",&K)!=EOF){
        getchar() ;
        gets(old_str) ;
        Len=0 ;
        L_old=strlen(old_str) ;
        for(int i=0;i<L_old;i++){
            if('A'<=old_str[i]&&old_str[i]<='Z'){
                my_hash[Len]=i+1 ;
                str[Len++]=old_str[i]-'A'+'a' ;
            }
            else if('a'<=old_str[i]&&old_str[i]<='z'){
                my_hash[Len]=i+1 ;
                str[Len++]=old_str[i] ;
            }
            else
                continue ;
        }
        str[Len]='\0' ;
        ans_len=0 ;
        for(int i=0;i<Len;i++){
            Node now=gao_case1(i) ;
            if(now.Lenght>ans_len){
                 ans_len=now.Lenght ;
                 ans_left_id=now.Left_id ;
            }
            else if(now.Lenght==ans_len)
                 ans_left_id=Min(ans_left_id,now.Left_id) ;
            now=gao_case2(i) ;
            if(now.Lenght>ans_len){
                 ans_len=now.Lenght ;
                 ans_left_id=now.Left_id ;
            }
            else if(now.Lenght==ans_len)
                 ans_left_id=Min(ans_left_id,now.Left_id) ;
        }
        printf("Case %d: %d %d\n",T++ ,ans_len,ans_left_id) ;
   }
   return 0 ;
}