bzoj 2251 [2010Beijing Wc]外星联络 后缀数组

题面

题目传送门

解法

用后缀数组求出height数组，然后暴力即可

时间复杂度：\(O(n^2)\)

代码

#include <bits/stdc++.h>
#define N 3010
using namespace std;
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
int n, st[N];
struct SuffixArray {
	int x[N], y[N], sa[N], rnk[N], cnt[N], height[N];
	void build() {
		int m = 2;
		for (int i = 1; i <= n; i++) cnt[x[i] = st[i]]++;
		for (int i = 2; i <= m; i++) cnt[i] += cnt[i - 1];
		for (int i = n; i; i--) sa[cnt[x[i]]--] = i;
		for (int k = 1; k <= n; k <<= 1) {
			int num = 0;
			for (int i = n - k + 1; i <= n; i++) y[++num] = i;
			for (int i = 1; i <= n; i++)
				if (sa[i] > k) y[++num] = sa[i] - k;
			for (int i = 1; i <= m; i++) cnt[i] = 0;
			for (int i = 1; i <= n; i++) cnt[x[i]]++;
			for (int i = 2; i <= m; i++) cnt[i] += cnt[i - 1];
			for (int i = n; i; i--) sa[cnt[x[y[i]]]--] = y[i], y[i] = 0;
			swap(x, y); x[sa[1]] = 1, num = 1;
			for (int i = 2; i <= n; i++)
				x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) ? num : ++num;
			if (num == n) break; m = num;
		}
		for (int i = 1; i <= n; i++) rnk[sa[i]] = i;
		int k = 0;
		for (int i = 1; i <= n; i++) {
			if (rnk[i] == 1) continue;
			if (k) k--; int j = sa[rnk[i] - 1];
			while (i + k <= n && j + k <= n && st[i + k] == st[j + k]) k++;
			height[rnk[i]] = k;
		}
	}
	void query() {
		int mx = 0;
		for (int i = 2; i <= n; i++) {
			if (height[i] > height[i - 1]) {
				for (int j = mx + 1; j <= height[i]; j++) {
					int k = i, tot = 1;
					while (height[k] >= j) k++, tot++;
					cout << tot << "\n";
				}
			}
			mx = height[i];
		}
	}
} SA;
int main() {
	read(n);
	for (int i = 1; i <= n; i++) {
		char c = getchar();
		while (!isdigit(c)) c = getchar();
		st[i] = c - '0' + 1;
	}
	SA.build(); SA.query();
	return 0;
}