最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 26222    Accepted Submission(s): 9649

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
 

Recommend
lcy 

#include<bits/stdc++.h>
#define FAST ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N=1e6+5;
const int MOD=1e9+7;
const int INF=0x3f3f3f3f;
char s[MAX_N];
char temp[MAX_N<<1];
int len[MAX_N<<1];

void init(){
    int l=strlen(s);
    temp[0]='@';
    for(int i=1;i<=2*l;i+=2)
        temp[i]='#',temp[i+1]=s[i/2];
    temp[2*l+1]='#';
    temp[2*l+2]='$';
    temp[2*l+3]='0';
}

int manacher(int l){
    int mx=0,ans=0,pos=0;
    for(int i=1;i<=l;++i){
        if(mx>i)    len[i]=min(mx-i,len[2*pos-i]);
        else    len[i]=1;
        while(temp[i-len[i]]==temp[i+len[i]])   len[i]++;
        if(len[i]+i>mx) mx=len[i]+i,pos=i;
        ans=max(ans,len[i]-1);
    }
    return ans;
}

int main(void){
    while((scanf("%s",s))!=EOF){
        int len=(int)strlen(s);
        init();
        int ans=manacher(len*2+1);
        cout <<ans <<endl;
    }

}