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 ;

}

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hdu3068

最长回文