[USACO06DEC]牛奶模式Milk Patterns - 后缀数组

Description

求字符串 \(s\) 的最长可重叠重复 \(k\) 次子串。

Solution

重复出现了 \(k\) 次，相当于我们选择了 \(k\) 个后缀，求他们的 \(lcp\)。

显然 \(k\) 个后缀的 \(rank\) 是连续的，所以重复出现 \(k\) 次的前缀就是 \(min(height[l+1\dots l+k-1])\)。

所以我们需枚举 \(i\)，然后用一个递增的单调队列维护 \(height[i-k+2\dots i]\) 即可。

Code

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

inline int ty() {
	char ch = getchar(); int x = 0, f = 1;
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}

const int _ = 2e4 + 10;
int N, K, s[_], ans = 0;
int rnk[_], sa[_], height[_];

void SA() {
	static int t[_], a[_], buc[_], fir[_], sec[_], tmp[_];
	copy(s + 1, s + N + 1, t + 1);
	sort(t + 1, t + N + 1);
	int *end = unique(t + 1, t + N + 1);
	for (rg int i = 1; i <= N; ++i) a[i] = lower_bound(t + 1, end, s[i]) - t;
	fill(buc + 1, buc + N + 1, 0);
	for (rg int i = 1; i <= N; ++i) ++buc[a[i]];
	for (rg int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
	for (rg int i = 1; i <= N; ++i) rnk[i] = buc[a[i] - 1] + 1;
	for (rg int len = 1; len <= N; len <<= 1) {
		for (rg int i = 1; i <= N; ++i) {
			fir[i] = rnk[i];
			sec[i] = i + len > N ? 0 : rnk[i + len];
		}
		fill(buc + 1, buc + N + 1, 0);
		for (rg int i = 1; i <= N; ++i) ++buc[sec[i]];
		for (rg int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
		for (rg int i = 1; i <= N; ++i) tmp[N - --buc[sec[i]]] = i;
		fill(buc + 1, buc + N + 1, 0);
		for (rg int i = 1; i <= N; ++i) ++buc[fir[i]];
		for (rg int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
		for (rg int i, j = 1; j <= N; ++j) {
			i = tmp[j];
			sa[buc[fir[i]]--] = i;
		}
		bool same = false;
		for (int i, j = 1, last = 0; j <= N; ++j) {
			i = sa[j];
			if (!last) rnk[i] = 1;
			else if (fir[i] == fir[last] && sec[i] == sec[last]) rnk[i] = rnk[last], same = true;
			else rnk[i] = rnk[last] + 1;
			last = i;
		}
		if (!same) break;
	}
  for (int i = 1, k = 0; i <= N; ++i) {
    if (rnk[i] == 1)
      k = 0;
    else {
      if (k > 0) --k;
      int j = sa[rnk[i] - 1];
      while (i + k <= N && j + k <= N && a[i + k] == a[j + k]) ++k;
    }
    height[rnk[i]] = k;
  }
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("milk.in", "r", stdin);
	freopen("milk.out", "w", stdout);
#endif
	N = ty(), K = ty();
	for (rg int i = 1; i <= N; ++i) s[i] = ty();
	SA();
	/*for (int i = 1; i <= N; ++i) printf("%d ", height[i]);
	puts("");*/
	static int q[_], l, r;
	q[l = r = 1] = 0;
	for (int i = 1; i <= N; ++i) {
		while (l < r && q[l] <= i - K + 1) ++l;
		while (l < r && height[q[r - 1]] >= height[i]) --r;
		q[r++] = i;
		if (i >= K) ans = max(ans, height[q[l]]);
	}
	printf("%d\n", ans);
	return 0;
}