hdoj 3068最长回文【Manacher】

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12270    Accepted Submission(s): 4511

Problem Description

给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度.
回文就是正反读都是一样的字符串,如aba, abba等

 

Input

输入有多组case,不超过120组,每组输入为一行小写英文字符a,b,c...y,z组成的字符串S
两组case之间由空行隔开(该空行不用处理)
字符串长度len <= 110000

 

Output

每一行一个整数x,对应一组case,表示该组case的字符串中所包含的最长回文长度.

 

Sample Input

aaaaabab

 

Sample Output

43

 

Source

2009 Multi-University Training Contest 16 - Host by NIT 

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int N = 110005;char s[N],a[N*2];int p[N*2];int len;void manacher(char *s){int l = 0;a[l++] = '$';a[l++] = '#';for(int i = 0; i < len; i++){a[l++] = s[i];a[l++] = '#';}a[l] = 0;int id = 0, ml = 0;for(int i = 0; i < 2*len+2; i++){if(p[id]+id > i)p[i] = min(p[2*id-i], p[id]+id-i);elsep[i] = 1;while(a[i+p[i]] == a[i-p[i]])p[i]++;if(id+p[id] < i+p[i])id = i;if(ml < p[i])ml = p[i];}printf("%d\n", ml-1);} int main(){while(scanf("%s",s)!=EOF){len = strlen(s);manacher(s);}return 0; }