HDU 3068 最长回文 [Manacher回文串]

最长回文
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17805 Accepted Submission(s): 6568

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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Manacher裸题嘛。
重点在于分成两类考虑。
第一种是之前最大回文串能够到达的点在当前点左边，那么这时候默认p[i]=1，因为单独一个也是回文串，然后再暴力扩展。
第二种的话，找到对称点然后判断对称点的p[id*2-i]与当前id及p[id]所允许的最大宽度，最后同样对于恰好就在最大宽度处的情况做一次暴力扩展。
另外最后注意ans = maxlen - 1

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<climits>#include<cctype>#include<algorithm>#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std;#define smax(x,tmp) x=max((x),(tmp))#define smin(x,tmp) x=min((x),(tmp))#define maxx(x1,x2,x3) max(max(x1,x2),x3)#define minn(x1,x2,x3) min(min(x1,x2),x3)const int INF=0x3f3f3f3f;const int maxn = 220010;char s[maxn],cpy[maxn];int p[maxn];int lens;int n;inline int manacher(){    memset(p,0,sizeof(p));    for(int i=lens;i>=0;i--) // start from s[lens]='\0'    {        s[(i<<1|1)+1]=s[i];        s[(i<<1|1)]='#';    }    s[0]='$';    int id = 0 , maxlen = 0;    for(int i=2;i<(lens<<1|1);i++)    {        if(p[id]+id>i) p[i] = min(p[id*2-i],p[id]-(i-id));        else p[i]=1;        while(s[i-p[i]] == s[i+p[i]]) ++p[i];        if(id+p[id] < i+p[i]) id=i;        smax(maxlen,p[i]);    }    return maxlen-1;}int main(){#ifndef ONLINE_JUDGE    freopen("rev.in","r",stdin);    freopen("rev.out","w",stdout);#endif    while(~scanf("%s",s) && (lens = strlen(s)))    {        int ans = manacher();        printf("%d\n",ans);    }    return 0;}