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P4070 [SDOI2016] 生成魔咒

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P4070 [SDOI2016] 生成魔咒

题面

题目链接

解法

注意到每次加入一个字符$s[i]$后，增加的不同子串数量就是$i-lcs$，其中$lcs$是子串$s[1\cdots i]$和加入前$s$中任意子串最长的最长公共后缀长度。

要求出每个前缀和其他子串的lcs，容易想到将串反转后使用后缀数组。现在问题转化为对每个$i$求出对应的$p \lt i$使得$\min\limits_{k\in(rk[p],rk[i]]\bigcup(rk[i],rk[p]]}{height[k]}$最大，显然这个$rk[p]$是$rk[i]$的前驱或者后继。

按顺序枚举$i$，用平衡树两棵线段树维护前驱后继，求出lcs取max，得到每次增加的不同子串数量，得到最终答案。

时间复杂度$O(n \log n)$。

AC代码

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/**
 * @file:           P4070.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P4070
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp& x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
const int MAXN = 1e5 + 10;
const int LOGN = 19;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct SuffixArray {
    int n, sa[MAXN], rk[MAXN], tp[MAXN], ht[MAXN], he[MAXN];
    void radix_sort(int m) {
        static int buc[MAXN];
        for (int i = 0; i <= m; ++i)
            buc[i] = 0;
        for (int i = 1; i <= n; ++i)
            buc[rk[i]]++;
        for (int i = 1; i <= m; ++i)
            buc[i] += buc[i - 1];
        for (int i = n; i >= 1; --i)
            sa[buc[rk[tp[i]]]--] = tp[i];
    }
    void init(int n, int* s) {
        this->n = n;
        int m = 1e5;
        for (int i = 1; i <= n; ++i)
            rk[i] = s[i], tp[i] = i;
        radix_sort(m);
        for (int w = 1, p = 0; p < n; m = p, w <<= 1) {
            p = 0;
            for (int i = 1; i <= w; ++i)
                tp[++p] = n - w + i;
            for (int i = 1; i <= n; ++i)
                if (sa[i] > w)
                    tp[++p] = sa[i] - w;
            radix_sort(m);
            copy(rk + 1, rk + n + 1, tp + 1);
            rk[sa[1]] = p = 1;
            for (int i = 2; i <= n; ++i) {
                if (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])
                    rk[sa[i]] = p;
                else
                    rk[sa[i]] = ++p;
            }
        }
        for (int i = 1, k = 0; i <= n; ++i) {
            if (k)
                k--;
            while (s[i + k] == s[sa[rk[i] - 1] + k])
                k++;
            ht[i] = k;
        }
        for (int i = 1; i <= n; ++i)
            he[i] = ht[sa[i]];
    }
} sa;
#define li (i << 1)
#define ri (i << 1) | 1
#define lson li, l, mid
#define rson ri, mid + 1, r
template <class _Update>
struct SegmentTree {
    _Update upd;
    int nd[MAXN * 4];
    void build(int i, int l, int r, int x) {
        nd[i] = x;
        if (l == r)
            return;
        int mid = (l + r) >> 1;
        build(lson, x), build(rson, x);
    }
    void update(int i, int l, int r, int q, int v) {
        if (l == r)
            return void(nd[i] = upd(nd[i], v));
        int mid = (l + r) >> 1;
        if (q <= mid)
            update(lson, q, v);
        else
            update(rson, q, v);
        nd[i] = upd(nd[li], nd[ri]);
    }
    int query(int i, int l, int r, int ql, int qr, int x) {
        if (ql > qr)
            return x;
        if (ql <= l && r <= qr)
            return nd[i];
        int mid = (l + r) >> 1, ret = x;
        if (ql <= mid)
            ret = upd(ret, query(lson, ql, qr, x));
        if (qr > mid)
            ret = upd(ret, query(rson, ql, qr, x));
        return ret;
    }
};
struct MaxUpdate { int operator()(int x, int y) { return max(x, y); } };
struct MinUpdate { int operator()(int x, int y) { return min(x, y); } };
int n, m;
int a[MAXN], b[MAXN], rev[MAXN], lg2[MAXN], st[MAXN][LOGN];
SegmentTree<MaxUpdate> sgtpre;
SegmentTree<MinUpdate> sgtnxt;
int lcs(int l, int r) {
    if (l == r)
        return INF;
    int k = lg2[r - l];
    return min(st[l + 1][k], st[r - (1 << k) + 1][k]);
}
signed main() {
    read(n), m = n;
    for (int i = 1; i <= n; ++i)
        read(a[i]), b[i] = a[i];
    sort(b + 1, b + m + 1);
    m = unique(b + 1, b + m + 1) - b - 1;
    for (int i = 1; i <= n; ++i)
        a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
    copy(a + 1, a + n + 1, rev + 1);
    reverse(rev + 1, rev + n + 1);
    sa.init(n, rev);
    for (int i = 2; i <= n; ++i)
        lg2[i] = (i & (i - 1)) ? lg2[i - 1] : lg2[i - 1] + 1;
    for (int i = 1; i <= n; ++i)
        st[i][0] = sa.he[i];
    for (int j = 1; j < LOGN; ++j) {
        for (int i = 1; i + (1 << j) - 1 <= n; ++i) {
            st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
        }
    }
    sgtpre.build(1, 1, n, 0);
    sgtnxt.build(1, 1, n, n + 1);
    int cur = 0;
    for (int i = 1; i <= n; ++i) {
        int t = sa.rk[n - i + 1];
        int p1 = sgtpre.query(1, 1, n, 1, t, 0);
        int p2 = sgtnxt.query(1, 1, n, t, n, n + 1);
        cur += i - max(lcs(p1, t), lcs(t, p2));
        sgtpre.update(1, 1, n, t, t);
        sgtnxt.update(1, 1, n, t, t);
        write(cur), putchar('\n');
    }
    return 0;
}