[回文自动机] BZOJ 3676: [Apio2014]回文串

最新推荐文章于 2019-02-22 14:16:00 发布

原创最新推荐文章于 2019-02-22 14:16:00 发布 · 384 阅读

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本文介绍了一种基于回文自动机的算法实现，通过在每个节点维护一个right值来求解最大回文子串的问题。该算法类似于后缀自动机，并通过势能分析讨论了其复杂度。

$Solution$

回文自动机有很多性质是和后缀自动机很相似的。
这个题可以类似于SAM，在每个节点 $u$ 维护一个 $right_u$ 。答案就是 $max\{|right_u|*len_u\}$
关于回文自动机的复杂度，可以势能分析。

#include <bits/stdc++.h>
using namespace std;

const int N = 303030;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}

int ch[N][30];
int par[N], ri[N], len[N];
char s[N];
int last, Tcnt;
int n;
long long ans;

inline int Extend(int key, int pos) {
    int p = last, np, q;
    while (s[pos] != s[pos - len[p] - 1]) p = par[p];
    if (!ch[p][key]) {
        np = ++Tcnt; q = par[p];
        len[np] = len[p] + 2;
        while (s[pos] != s[pos - len[q] - 1]) q = par[q];
        par[np] = ch[q][key]; ch[p][key] = np;
    }
    np = ch[p][key];
    ++ri[np]; return last = np;
}

int main(void) {
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    scanf("%s", s + 1);
    n = strlen(s + 1);
    par[1] = par[0] = 1;
    len[1] = -1; Tcnt = 1;
    for (int i = 1; i <= n; i++) Extend(s[i] - 'a', i);
    for (int i = Tcnt; ~i; i--) {
        ri[par[i]] += ri[i];
        ans = max(ans, (long long)ri[i] * len[i]);
    }
    cout << ans << endl;
    return 0;
}

UPD：
后来用论文里不基于势能分析的插入算法写了一发。
记录一个 $quick_u$ 直接找到 $fail$ 链上的合法点。
多了一个 $O(\Sigma)$ 的复杂度。

#include <bits/stdc++.h>
using namespace std;

const int N = 303030;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}

int ch[N][30], qui[N][30];
int par[N];
int ri[N], len[N];
char s[N];
int last, Tcnt;
int n;
long long ans;

inline int Extend(int pos) {
    int p = last, np, q, key = s[pos];
    if (s[pos - len[p] - 1] != key) p = qui[p][key];
    if (!ch[p][key]) {
        np = ++Tcnt; q = par[p];
        len[np] = len[p] + 2;
        if (s[pos - len[q] - 1] == key) par[np] = ch[q][key];
        else par[np] = ch[qui[q][key]][key];
        memcpy(qui[np], qui[par[np]], sizeof qui[np]);
        qui[np][s[pos - len[par[np]]]] = par[np];
        ch[p][key] = np;
    }
    np = ch[p][key];
    ++ri[np]; return last = np;
}

int main(void) {
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    scanf("%s", s + 1);
    n = strlen(s + 1);
    par[0] = par[1] = 1;
    for (int i = 0; i < 26; i++)
        qui[0][i] = qui[1][i] = 1;
    len[1] = -1; last = Tcnt = 1;
    s[0] = -1;
    for (int i = 1; i <= n; i++) {
        s[i] -= 'a'; Extend(i);
    }
    for (int i = Tcnt; ~i; i--) {
        ri[par[i]] += ri[i];
        ans = max(ans, (long long)ri[i] * len[i]);
    }
    cout << ans << endl;
    return 0;
}