题目链接

题目描述

给定一个字符串，求至少出现 \(k\) 次的最长重复子串，这 \(k\) 个子串可以重叠。

思路

二分 子串长度，据其将 \(h\) 数组 分组，判断是否存在一组其大小 \(\geq k\).

Code

#include <cstdio>

#include <vector>

#include <iostream>

#define maxn 1000000

#define maxm maxn + 10

#define maxd 20010

using namespace std;

typedef long long LL;

int wa[maxm], wb[maxm], wv[maxm], wt[maxm], r[maxd], rk[maxd], sa[maxd], n, m, k, h[maxd], a[maxn];

vector<int> v[maxd];

bool cmp(int* r, int a, int b, int l) { return r[a] == r[b] && r[a+l] == r[b+l]; }

void init(int* r, int* sa, int n, int m) {

    int* x=wa, *y=wb, *t, i, j, p;

    for (i = 0; i < m; ++i) wt[i] = 0;

    for (i = 0; i < n; ++i) ++wt[x[i] = r[i]];

    for (i = 1; i < m; ++i) wt[i] += wt[i - 1];

    for (i = n-1; i >= 0; --i) sa[--wt[x[i]]] = i;

    for (j = 1; p < n; j <<= 1, m = p) {

        for (p = 0, i = n-j; i < n; ++i) y[p++] = i;

        for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;

        for (i = 0; i < n; ++i) wv[i] = x[y[i]];

        for (i = 0; i < m; ++i) wt[i] = 0;

        for (i = 0; i < n; ++i) ++wt[wv[i]];

        for (i = 1; i < m; ++i) wt[i] += wt[i - 1];

        for (i = n-1; i >= 0; --i) sa[--wt[wv[i]]] = y[i];

        t = x, x = y, y = t, x[sa[0]] = 0;

        for (p = 1, i = 1; i < n; ++i) x[sa[i]] = cmp(y, sa[i], sa[i-1], j) ? p - 1 : p++;

    }

    for (i = 0; i < n; ++i) rk[sa[i]] = i;

    int k = 0;

    for (i = 0; i < n; h[rk[i++]] = k) {

        for (k = k ? --k : 0, j = sa[rk[i] - 1]; r[i+k] == r[j+k]; ++k);

    }

}

bool check(int x) {

    int cnt = -1;

    for (int i = 1; i < n; ++i) {

        if (h[i] < x) v[++cnt].clear();

        v[cnt].push_back(h[i]);

    }

    for (int i = 0; i <= cnt; ++i) if (v[i].size() >= k) return true;

    return false;

}

int main() {

    int m = 0;

    scanf("%d%d", &n, &k);

    for (int i = 0; i < n; ++i) {

        scanf("%d", &a[i]);

        m = max(m, a[i]);

    }

    r[n++] = 0;

    init(a, sa, n, ++m);

    int le = 1, ri = n - 1;

    while (ri - le > 1) {

        int mid = le + ri >> 1;

        if (check(mid)) le = mid;

        else ri = mid - 1;

    }

    if (check(ri)) printf("%d\n", ri);

    else printf("%d\n", le);

    return 0;

}