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P1117 [NOI2016] 优秀的拆分

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P1117 [NOI2016] 优秀的拆分

题面

题目链接

解法

$O(n^2)$暴力+$hash$能拿$95pts$。

但显然我们不满足于这$95pts$，考虑如何优化这个$O(n^2)$，即更快地求出每个位置对应的AA串数量。

考虑枚举A串的长度$len$，并且每隔$len$个字符分割一次，这样每一条长度为$len\times 2$的AA串都经过了至少$2$个分隔符，于是考虑从分隔符向两边拓展即可，用后缀数组计算。时间复杂度$O(n\log n)$。

鬼知道我调了多久（具体看代码吧）。SA多次使用一定要清空，我也不知道为什么。

AC代码

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/**
 * @file:           P1117.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P1117
*/
// #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 = 3e4 + 10;
const int LOGN = 16;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct SuffixArray {
    int n, sa[MAXN], rk[MAXN], tp[MAXN], he[MAXN];
    void clear() {
        fill(sa, sa + n + 1, 0);
        fill(rk, rk + n + 1, 0);
        fill(tp, tp + n + 1, 0);
        fill(he, he + n + 1, 0);
        n = 0;
    }
    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, char* s) {
        this->n = n;
        int m = 200;
        for (int i = 1; i <= n; ++i)
            rk[i] = s[i] + 1, 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++;
            he[rk[i]] = k;
        }
    }
} sa, pa;
int n, a[MAXN], b[MAXN], lg2[MAXN];
int smn[MAXN][LOGN], pmn[MAXN][LOGN];
char s[MAXN];
int getlcs(int l, int r) {
    if (l > r)
        swap(l, r);
    int k = lg2[r - l];
    return min(smn[l + 1][k], smn[r - (1 << k) + 1][k]);
}
int getlcp(int l, int r) {
    if (l > r)
        swap(l, r);
    int k = lg2[r - l];
    return min(pmn[l + 1][k], pmn[r - (1 << k) + 1][k]);
}
signed main() {
    int tt;
    read(tt);
    while (tt--) {
        scanf("%s", s + 1);
        n = strlen(s + 1);
        sa.init(n, s);
        reverse(s + 1, s + n + 1);
        pa.init(n, s);
        reverse(s + 1, s + n + 1);
        fill(a, a + n + 2, 0);
        fill(b, b + n + 2, 0);
        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) {
            smn[i][0] = sa.he[i];
            pmn[i][0] = pa.he[i];
        }
        for (int j = 1; j < LOGN; ++j) {
            for (int i = 1; i + (1 << j) - 1 <= n; ++i) {
                smn[i][j] = min(smn[i][j - 1], smn[i + (1 << (j - 1))][j - 1]);
                pmn[i][j] = min(pmn[i][j - 1], pmn[i + (1 << (j - 1))][j - 1]);
            }
        }
        for (int x = 1; x + x <= n; ++x) {
            for (int sr = x + 1; sr <= n; sr += x) {
                int sl = sr - x;
                int pl = n - sr + 1, pr = n - sl + 1;
                int pre = getlcp(pa.rk[pl], pa.rk[pr]);
                int nxt = getlcs(sa.rk[sl], sa.rk[sr]);
                pre = min(pre, x), nxt = min(nxt, x);
                if (pre + nxt > x) {
                    a[sr - pre + x]++, a[sr + nxt]--;
                    b[sl - pre]--, b[sl + nxt - x]++;
                }
            }
        }
        for (int i = 1; i <= n; ++i)
            a[i] += a[i - 1];
        for (int i = n; i >= 1; --i)
            b[i] += b[i + 1];
        int ans = 0;
        for (int i = 1; i < n; ++i)
            ans += a[i] * b[i + 1];
        write(ans), putchar('\n');
        sa.clear(), pa.clear();
    }
    return 0;
}