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Manacher(输出最长回文串及下标)

时间:2019-08-03 21:48:13      阅读:114      评论:0      收藏:0      [点我收藏+]

标签:only   ever   inpu   ane   content   comm   fine   abc   sep   

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
 

 

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 ;
}

 

Manacher(输出最长回文串及下标)

标签:only   ever   inpu   ane   content   comm   fine   abc   sep   

原文地址:https://www.cnblogs.com/nonames/p/11296274.html

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