回文树

例题

PAM

用以处理回文串问题的一类自动机
每个节点代表一类回文串

节点信息：

回文串长度，fail指针，子节点，出现次数等等

初始化

初始化回文串的时，建立两个节点，长度分别为\(-1\)和\(0\)，代表奇数回文串和偶数回文串，并标记偶数节点的\(fail\)为奇数节点【当任意长度的回文串都不存在时，那么节点自身就作为它结尾的最长回文串】
同时记一个\(last\)为上一个插入的节点

PAM(){len[siz = 1] = -1,s[0] = -1,fail[0] = 1;}

插入操作

不断沿\(last\)的\(fail\)指针往上找，直至找到\(s[n - 1 - len[u]] = s[n]\)的节点\(u\)，便可插入其后
设置\(fail\)指针时，同样沿\(fail\)指针查找，直至找到\(s[n - 1 - len[u]] = s[n]\)的节点\(u\)，作为\(fail\)

int find(int u){
    while (s[n - 1 - len[u]] != s[n]) u = fail[u];
    return u;
}
void  ins(int x){
    s[++n] = x; int cur = find(last);
    if (!ch[cur][x]){
        fail[++siz] = ch[find(fail[cur])][x];
        ch[cur][x] = siz; len[siz] = len[cur] + 2;
    }
    last = ch[cur][x]; cnt[last]++;
}

统计

每个节点出现次数还与其儿子节点有关，所以统计时还要加上子树的贡献

LL count(){
    LL re = 0;
    for (int i = siz; ~i; i--){
        cnt[fail[i]] += cnt[i];
        re = max(re,1ll * cnt[i] * len[i]);
    }
    return re;
}

这样就可以\(A\)掉这道题辣

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 300005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
    return flag ? out : -out;
}
char s[maxn];
int n;
struct PAM{
    int n,last,siz,ch[maxn][26],fail[maxn],cnt[maxn],len[maxn],s[maxn];
    PAM(){len[siz = 1] = -1,s[0] = -1,fail[0] = 1;}
    int find(int u){
        while (s[n - 1 - len[u]] != s[n]) u = fail[u];
        return u;
    }
    void  ins(int x){
        s[++n] = x; int cur = find(last);
        if (!ch[cur][x]){
            fail[++siz] = ch[find(fail[cur])][x];
            ch[cur][x] = siz; len[siz] = len[cur] + 2;
        }
        last = ch[cur][x]; cnt[last]++;
    }
    LL count(){
        LL re = 0;
        for (int i = siz; ~i; i--){
            cnt[fail[i]] += cnt[i];
            re = max(re,1ll * cnt[i] * len[i]);
        }
        return re;
    }
}T;
int main(){
    scanf("%s",s + 1); n = strlen(s + 1);
    REP(i,n) T.ins(s[i] - 'a');
    printf("%lld\n",T.count());
    return 0;
}