hdu 3294 Girls' research ( manacher+特殊输出格式 )

最新推荐文章于 2022-04-30 19:23:50 发布

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最新推荐文章于 2022-04-30 19:23:50 发布

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文章标签： C++ 字符串 manacher

本文链接：https://blog.youkuaiyun.com/qq_24451605/article/details/44243511

本文介绍了一个关于寻找字符串中最长回文子串的问题，并提供了一种通过转换和Manacher算法来解决该问题的方法。具体步骤包括对输入字符串进行字符替换，然后使用Manacher算法找出最长回文子串。

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Girls' research

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 592 Accepted Submission(s): 224

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
题目大意:求最大回文子串,有输出要求
题目分析;就是模拟那部分麻烦点

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cctype>
#define MAX 200007

using namespace std;

char mp[507];
char s[MAX];
char temp[MAX<<1];
int Len[MAX<<1];
char ch[5];

int init ( char *st )
{
    int i , len = strlen ( st );
    temp[0] ='@';
    for ( i = 1 ; i <= 2*len ; i += 2 )
    {
        temp[i] = '#';
        temp[i+1] = st[i/2];
    }
    temp[2*len+1] = '#';
    temp[2*len+2] = '$';
    temp[2*len+3] = 0;
   // cout << "string : " << temp << endl;
    return 2*len+1;
}

int manacher ( char *st , int len ) 
{
    int mx = 0 , po = 0 , ans = 0;
    for ( int i = 1 ; i <= len ; i++ )
    {
        if ( mx > i )
            Len[i] = min ( mx - i , Len[2*po-i] );
        else 
            Len[i] = 1;
        while ( st[i-Len[i]] == st[i+Len[i]] )
            Len[i]++;
        if ( Len[i]+i > mx )
            mx = Len[i]+i , po = i;
        ans = max ( ans , Len[i] );
    }
    return ans;
}

int main ( )
{
    while ( ~scanf ( "%s" , ch ) )
    {
        scanf ( "%s" , s );
        for ( int i = 0 ; i < 26 ; i++ )
            mp[(ch[0]-'a'+i)%26] = 'a'+i;
        int len = strlen(s);
        for ( int i = 0 ; i < len ; i++ )
            s[i] = mp[s[i]-'a'];
       /*
        cout <<"----------mp------------" << endl;
        for ( int i = 0 ; i < 26 ; i++ )
            printf ( "%c " , mp[i] );
        puts("");
        cout << "-----------------------" << endl;
        */
        len = init(s);
        int mx = manacher ( temp , len );
        if ( mx <= 3 )
        {
            puts ( "No solution!" );
            continue;
        }
        int l , r;
        for ( int i = 1 ; i <= len ; i++ )
        {
            if ( Len[i] == mx ) 
            {
                l = (i-Len[i]+2)/2-1;
                r = (i+Len[i]-2)/2-1;
                printf ( "%d %d\n" , l , r );
                for ( int i = l ; i <= r ; i++ )
                printf ( "%c" , s[i] );
                puts ( "" );
                break;
            }
        }
        
    }
}