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最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9418    Accepted Submission(s): 3238

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4
3

原字符串的回文串的长度 == p[i]-1

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define maxn 310000
int p[maxn] ;
char s[maxn] , str[maxn] ;
int init() {
    int i = 0 , j = 0 , l = strlen(s) ;
    str[j++] = '$' ;
    while( i < l ) {
        str[j++] = '#' ;
        str[j++] = s[i] ;
        i++ ;
    }
    str[j++] = '#' ;
    str[j] = '\0' ;
    return j ;
}
void Manacer(int l) {
    int i , max1 = 0 , id ;
    for(i = 1 ; i < l ; i++) {
        if( max1 > i )
            p[i] = min(p[2*id-i],max1-i) ;
        else
            p[i] = 1 ;
        while( str[i-p[i]] == str[i+p[i]] )
            p[i]++ ;
        if( p[i]+i > max1) {
            max1 = p[i] + i ;
            id = i ;
        }
    }
}
int main() {
    int max1 , i , l ;
    while( scanf("%s", s) != EOF ) {
        l = init() ;
        Manacer(l) ;
        max1 = 0 ;
        for(i = 1 ; i < l ; i++) {
            max1 = max(max1,p[i]) ;
        }
        printf("%d\n", max1-1) ;
    }
    return 0 ;
}

hdu3068--最长回文(Manacer算法)

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原文地址：http://blog.csdn.net/winddreams/article/details/44218843