【BZOJ1396】【BZOJ2865】识别子串（SAM，线段树）

Description

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Solution

识别子串显然就是后缀自动机上右端点集合大小为1的点而且右端点的位置等于maxlen。
我们对于每一个sz[u]=1的点，有两个更新的方向：
1. 1~len[u]-len[fa[u]] 更新成 len[u] - i + 1
2. len[u]-len[fa[u]]~len[u] 更新成 len[fa[u]] + 1
第二种直接用一棵线段树维护，第二种加上i后用线段树维护，最后加上i。
最后在两棵线段树上取min即可。

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Code

/**************************************
 * Au: Hany01
 * Prob: [BZOJ1396] 识别子串
 * Date: Jun 9th, 2018
 * Email: [email protected]
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#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 SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define y1 wozenmezhemecaia 
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

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 char c_; register int _, __;
    for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-')  __ = -1;
    for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 100005;

int n, tot = 1, ch[maxn << 1][26], fa[maxn << 1], sz[maxn << 1], len[maxn << 1], bkt[maxn], per[maxn << 1], las = 1;
char s[maxn];

inline void extend(int c, int id)
{
    int np = ++ tot, p = las;
    las = np, len[np] = len[p] + 1, sz[np] = 1;
    while (p && !ch[p][c]) ch[p][c] = np, p = fa[p];
    if (!p) fa[np] = 1;
    else {
        int q = ch[p][c];
        if (len[q] == len[p] + 1) fa[np] = q;
        else {
            int nq = ++ tot;
            len[nq] = len[p] + 1, Cpy(ch[nq], ch[q]), fa[nq] = fa[q], fa[q] = fa[np] = nq;
            while (p && ch[p][c] == q) ch[p][c] = nq, p = fa[p];
        }
    }
}

struct SegmentTree
{
    int tr[maxn << 2];

#define mid ((l + r) >> 1)
#define lc (t << 1)
#define rc (lc | 1)

    inline void init() { Set(tr, 127); }

    inline void pushdown(int t) { chkmin(tr[lc], tr[t]), chkmin(tr[rc], tr[t]); }

    inline void update(int t, int l, int r, int x, int y, int v)
    {
        if (x <= l && r <= y) { chkmin(tr[t], v); return; }
        pushdown(t);
        if (x <= mid) update(lc, l, mid, x, y, v);
        if (y > mid)  update(rc, mid + 1, r, x, y, v);
    }

    inline int query(int t, int l, int r, int x)
    {
        if (l == r) return tr[t];
        pushdown(t);
        if (x <= mid) return query(lc, l, mid, x);
        else return query(rc, mid + 1, r, x);
    }

}ST1, ST2;

int main()
{
#ifdef hany01
    File("bzoj1396");
#endif

    scanf("%s", s + 1);
    For(i, 1, n = strlen(s + 1)) extend(s[i] - 97, i);

    For(i, 1, tot) ++ bkt[len[i]];
    For(i, 1, n) bkt[i] += bkt[i - 1];
    For(i, 1, tot) per[bkt[len[i]] --] = i;
    ST1.init(), ST2.init();
    Fordown(i, tot, 2) {
        int u = per[i];
        sz[fa[u]] += sz[u];
        if (sz[u] == 1)
            ST1.update(1, 1, n, len[u] - len[u] + 1, len[u] - len[fa[u]], len[u] + len[u] - len[u] + 1),
            ST2.update(1, 1, n, len[u] - len[fa[u]], len[u], len[fa[u]] + 1);
    }

    For(i, 1, n) printf("%d\n", min(ST1.query(1, 1, n, i) - i, ST2.query(1, 1, n, i)));

    return 0;
}