HDU 3068 最长回文

最长回文

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

aaaa

abab

Sample Output

4

3

Source

2009 Multi-University Training Contest 16 - Host by NIT

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#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 20000050
char s[maxn],str[2*maxn];
int p[maxn];
void Manacher(int *p,char *str,int len){
int mx=0,idx=0;
for(int i=1;i<=len;i++){
p[i] = mx>i ? min(p[2*idx-i],mx-i) : 1;
while(str[i+p[i]]==str[i-p[i]]) p[i]++;
if(i+p[i] > mx){
mx=i+p[i];
idx=i;
}
}
}
int main(){
while(scanf("%s",s)!=EOF){
int nn=strlen(s);
int n=nn*2+2;
str[0]='$';
for(int i=0;i<=nn;i++){
str[i*2+1]='#';
str[i*2+2]=s[i];
}
Manacher(p,str,n);
int ans=1;
for(int i=0;i<n;i++) ans=max(ans,p[i]);
printf("%d\n",ans-1);
}
return 0;
}

Manacher算法裸题