【回文树】

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以HYSBZ 3676 回文串 为例

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int maxn = 3e5+5;
const int sgsz = 26;

//回文树
struct Pam
{
    int nxt[maxn][sgsz], S[maxn];
    int fail[maxn], num[maxn], cnt[maxn], len[maxn];
    int last, n, p;

    int Newnode(int l)
    {
        memset(nxt[p], 0, sizeof(nxt[p]));
        cnt[p] = num[p] = 0;
        len[p] = l;
        return p++;
    }

    void Init()
    {
        p = 0;
        Newnode(0);
        Newnode(-1);
        last = 0;
        n = 0;
        S[n] = -1;
        fail[0] = 1;
    }

    int Get_fail(int x)
    {
        while(S[n - len[x] - 1] != S[n])
            x = fail[x];
        return x;
    }

    void Add(int c)
    {
        c -= 'a';
        S[++n] = c;
        int cur = Get_fail(last);
        if(!nxt[cur][c])
        {
            int now = Newnode(len[cur] + 2);
            fail[now] = nxt[Get_fail(fail[cur])][c];
            nxt[cur][c] = now;
            num[now] = num[fail[now]] + 1;
        }
        last = nxt[cur][c];
        cnt[last]++;
    }

    void Count()
    {
        for(int i = p - 1; i >= 0; --i)
            cnt[fail[i]] += cnt[i];
    }

    LL Ans()
    {
        LL ans = 0;
        Count();
        for(int i = 0; i < p; i++)
            ans = max(ans, (LL)len[i] * (LL)cnt[i]);
        return ans;
    }

}pam;

int main()
{
    char s[maxn];
    scanf("%s", s);
    pam.Init();
    int l = strlen(s);
    for(int i = 0; i < l; ++i)
        pam.Add(s[i]);
    printf("%lld\n", pam.Ans());
    return 0;
}