Problem E : Find Palindrome


From: DHUOJ, 2017060305

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<center> Time Limit: 1 s </center>

Description

Given a string S, which consists of lowercase characters, you need to find the longest palindromic sub-string.

A sub-string of a string S  is another string S'  that occurs "in" S. For example, "abst" is a sub-string of "abaabsta". A palindrome is a sequence of characters which reads the same backward as forward.

 

Input

There are several test cases.
Each test case consists of one line with a single string S (1 ≤ |S | ≤ 50).

 

Output

For each test case, output the length of the longest palindromic sub-string.

 

Sample Input 

sasadasa
bxabx
zhuyuan

 

Sample Output 

7
1
3

 

Author: Kenny


利用dp来记录当前位置的最长回文串长度


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; 
#define MIN(a,b) ((a) < (b) ? (a) : (b))
  
int maxid;        // 最长回文子串下标
int LPS_rb[100];  // i为中心的回文子串右边界下标right border
char str[100];    // 原字符串处理后的副本
int maxlen;
void LPS_linear(char * X, int xlen)
{
    maxlen = maxid = 0;
  
    str[0] = '$';  // 将原串处理成所需的形式
    char *p = str;
    *(++p)++ = '#';
    while((*p++ = *X++) != '\0')
    {
        *p++ = '#';
    }
  
    for(int i = 1; str[i]; ++i)  // 计算LPS_rb的值
    {
        if(maxlen > i)          // 初始化LPS[i]
        {
            LPS_rb[i] = MIN(LPS_rb[2*maxid-i],(maxlen-i));
        }else
        {
            LPS_rb[i] = 1;
        }
        while(str[i-LPS_rb[i]] == str[i+LPS_rb[i]]) // 扩展
        {
            ++LPS_rb[i];
        }
        if(LPS_rb[i]-1 > maxlen)
        {
            maxlen = LPS_rb[i]-1;
            maxid = i;
        }
    }
}
 
int main(){
    char s[51];
    while(cin>>s){
        LPS_linear(s,strlen(s));
        printf("%d\n", maxlen);
    }
    return 0;
}