【BZOJ2342】【SHOI2011】双倍回文（回文树）

Description

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Solution

建出回文树，对fail树进行DFS，如果当前达到的点的len为4的倍数，且其祖先中有len是其一半的点，那么更新答案即可。

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Code

/************************************************
 * Au: Hany01
 * Date: Jun 22nd, 2018
 * Prob: [BZOJ2342][SHOI2011] 双倍回文
 * Email: [email protected]
 * Institute: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 500005;

struct PalindramicTree
{
    int ch[maxn][26], fa[maxn], len[maxn], tot, las, n, v[maxn], beg[maxn], nex[maxn], e, Ans, vis[maxn];
    char s[maxn];

    inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }

    inline void extend(int i)
    {
        int c = s[i] - 97, p = las;
        while (s[i - len[p] - 1] != s[i]) p = fa[p];
        if (!ch[p][c]) {
            int np = ++ tot, k = fa[p];
            len[np] = len[p] + 2;
            while (s[i - len[k] - 1] != s[i]) k = fa[k];
            fa[np] = ch[k][c], add(ch[k][c], np), ch[p][c] = np;
        }
        las = ch[p][c];
    }

    inline void init()
    {
        fa[0] = fa[1] = 1, add(1, 0), len[tot = 1] = -1;
        For(i, 1, n) extend(i);
    }

    void dfs(int u)
    {
        if (!(len[u] % 4) && vis[len[u] / 2]) chkmax(Ans, len[u]);
        if (u != 1) ++ vis[len[u]];
        for (register int i = beg[u]; i; i = nex[i]) dfs(v[i]);
        if (u != 1) -- vis[len[u]];
    }

}PT;

int main()
{
#ifdef hany01
    File("bzoj2342");
#endif

    PT.n = read(), scanf("%s", PT.s + 1);
    PT.init(), PT.dfs(1);
    printf("%d\n", PT.Ans);

    return 0;
}
//入我相思门，知我相思苦。
//    -- 李白《秋风词》