hdu3068

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17251    Accepted Submission(s): 6351

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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题解：

manacher算法的裸体

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#define N 2000100
int p[N];
char s[N],S[N];
int manacher(int len){
    int ans=,id=,mx=;
    for(int i=;i<len;i++){
        p[i]=mx>i?min(p[id+id-i],mx-i):;
        while(S[i-p[i]]==S[i+p[i]]) p[i]++;
        if(i+p[i]>mx) mx=i+p[i],id=i;
        ans=max(ans,p[i]-);
    }
    return ans;
}
void deal(int len){
    int l=;
    S[l++]='$';S[l++]='#';
    for(int i=;i<len;i++) S[l++]=s[i],S[l++]='#';
    S[l++]='\0';
}
int main(){
    //freopen("sh.txt","r",stdin);
    while(scanf("%s",s)==){
        memset(S,,sizeof S);
        memset(p,,sizeof p);
        int len=strlen(s);
        deal(len);
        printf("%d\n",manacher(len+len+));
        //fill(S,S+len+len+2,0);
        //fill(p,p+len+len+2,0);
    }
    return ;
}

A2：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#define N 210100
int p[N],l;
char s[N],S[N];
int manacher(){
    int ans=,mx=-,id=-;
    for(int i=;i<=l;i++){
        if(id+mx>=i) p[i]=min(p[(id<<)-i],id+mx-i);
        while(i-p[i]->=&&i+p[i]+<=l&&S[i-p[i]-]==S[i+p[i]+]) p[i]++;
        if(i+p[i]>id+mx) mx=p[i],id=i;
        ans=max(ans,p[i]);
    }
    return ans;
}
int main(){
    while(scanf("%s",s)==){
        memset(S,,sizeof S);
        memset(p,,sizeof p);
        int len=strlen(s);
        l=-;
        for(int i=;i<len;i++) S[++l]='#', S[++l]=s[i];
        S[++l]='#';
        printf("%d\n",manacher());
    }
    return ;
}