Manacher（输出最长回文串及下标）

Posted 2020-11-10 无名菜鸟1

http://acm.hdu.edu.cn/showproblem.php?pid=3294

Girls‘ research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 5711    Accepted Submission(s): 2117

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

 

 

Recommend

lcy

题意：先转换，再输出最长回文串及下标。

思路：利用原数组与加工后的数组的数量关系得到下标。

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
#define INF  10000000
using namespace std;
char a[200009] , b[400009] , str[200009];
int p[400009];


int num;


int main()
{
    char c ;
    while(~scanf("%c%s" , &c , a))
    {
        int len = strlen(a);
        int q = c - ‘a‘;
        memset(str , 0 , sizeof(str));
        memset(p , 0 , sizeof(p));
        for(int i = 0 ; i < len ; i++)
        {
            if(a[i] - q < ‘a‘)
            {
                a[i] = a[i] + 26;
            }
            a[i] = a[i] - q ;
        }
        //printf("%s\n" , a);
        int l = 0 ;
        b[l++] = ‘$‘;
        b[l++] = ‘#‘;
        for(int i = 0 ; i < len ; i++)
        {
            b[l++] = a[i];
            b[l++] = ‘#‘;
        }
        int mx = - 1 , mid , ans = 0;
        for(int i = 1 ; i < l ; i++)
        {
            if(mx > i)
            {
                p[i] = min(p[2*mid-i] , mx - i);
            }
            else
            {
                p[i] = 1 ;
            }
            while(b[i-p[i]] == b[i+p[i]])
            {
                p[i]++;
            }
            if(mx < p[i]+i)
            {
                mid = i ;
                mx = p[i] + i;
            }
        }
        int index = 0  , indey = 0;
        for(int i = 0 ; i < l ; i++)
        {
            if(ans < p[i] - 1)
            {
                ans = p[i] - 1;
                index = i ;

            }
        }
        if(ans == 1)
        {
            printf("No solution!\n");
        }
        else
        {
            int r = (index + ans)/2 - 1;
            int l = r - ans + 1;
            printf("%d %d\n" , l , r);
            for(int i = l ; i <= r ; i++)
            {
                printf("%c" , a[i]);
            }
            printf("\n");

        }
        getchar();
    }


    return 0 ;
}