后缀数组好题选讲(2)
涉及到的处理问题套路
用单调队列来处理长度为特定值的$LCP$出现次数
luogu P2852 Milk Pattern G
题意简述
给定一个长度为$n$的串，求出重复出现次数$\ge k$的子串数量
$n\le 2\cdot 10^4$
Sol
先将后缀排序，求出$height$数组
由于$height$数组代表的是$LCP(rnk[i],rnk[i-1])$，而这两个字符串开头所在的地方一定不同(毕竟是后缀数组)，那么$LCP[rnk[i],rnk[i-1]]$至少出现了$2$次
同理，区间$[i,i+k-1]$中$height$的最小值$s$代表某一个长度为$s$的子串至少出现了$k+1$次
所以一个长度为$k-1$的区间中$height$的最小值就是该区间中至少出现$k$次子串的最大长度
然后用单调队列解决即可。
#include <bits/stdc++.h> using namespace std; int n, k, m, num, a[50005], x[50005], y[50005], sa[50005], c[50005]; int rnk[50005], height[50005]; void get_SA() { m = 200; for (int i = 1; i <= n; i++) ++c[x[i] = a[i]]; for (int i = 2; i <= m; i++) c[i] += c[i - 1]; for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i; for (int k = 1; k <= n; k <<= 1) { 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; memset(c, 0, sizeof(c)); for (int i = 1; i <= n; i++) c[x[i]]++; for (int i = 2; i <= m; i++) c[i] += c[i - 1]; for (int i = n; i >= 1; i--) { sa[c[x[y[i]]]--] = y[i]; y[i] = 0; } swap(x, y); x[sa[1]] = 1; num = 1; for (int i = 2; i <= n; i++) { if (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num; else x[sa[i]] = ++num; } if (num == n) break; m = num; } } void get_height() { 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--; while (sa[rnk[i] - 1] + k <= n && i + k <= n && a[sa[rnk[i] - 1] + k] == a[i + k]) k++; height[rnk[i]] = k; } } int q[100005], pos[100005], hd = 1, tl = 0, maxn = 0; int main() { cin >> n >> k; k--; for (int i = 1; i <= n; i++) scanf("%d", &a[i]); get_SA(); get_height(); for (int i = 1; i <= n; i++) { while (hd <= tl && i - pos[hd] + 1 > k) ++hd; while (hd <= tl && height[i] <= q[tl]) --tl; q[++tl] = height[i]; pos[tl] = i; maxn = max(maxn, q[hd]); } cout << maxn << endl; }
luogu P5341 甲苯先生和大中锋的字符串
题意简述
给定一个长度为$n$的字符串，求出现次数恰好为$k$的字符串中出现次数最多的长度。
$n\le 10^5$
Sol
现在要求长度恰好为$k$，那么在我们的单调队列选定的$[i,i+k-2]$的长度为$k-1$的区间中，设其$height$最小值为$mn$，必须保证$\max (height[i-1],height[i+k-1]\le mn$，这样就满足了恰好$k$个的限制
#include <bits/stdc++.h> using namespace std; int n, m, k, num, x[100005], y[100005], c[100005], sa[100005], rnk[100005], height[100005]; char ch[500005]; void get_SA() { m = 200; for (int i = 1; i <= n; i++) ++c[x[i] = ch[i]]; for (int i = 2; i <= m; i++) c[i] += c[i - 1]; for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i; for (int k = 1; k <= n; k <<= 1) { 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; memset(c, 0, sizeof(c)); for (int i = 1; i <= n; i++) ++c[x[i]]; for (int i = 2; i <= m; i++) c[i] += c[i - 1]; for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i], y[i] = 0; swap(x, y); x[sa[1]] = 1; num = 1; for (int i = 2; i <= n; i++) if (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num; else x[sa[i]] = ++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 (j + k <= n && i + k <= n && ch[j + k] == ch[i + k]) k++; height[rnk[i]] = k; } } int q[500005], head = 1, tail = 0, cnt[500005]; int main() { int T; cin >> T; while (T--) { memset(x, 0, sizeof(x)); memset(y, 0, sizeof(y)); memset(rnk, 0, sizeof(rnk)); memset(c, 0, sizeof(c)); memset(sa, 0, sizeof(sa)); memset(height, 0, sizeof(height)); memset(cnt, 0, sizeof(cnt)); memset(q, 0, sizeof(q)); head = 1, tail = 0; scanf("%s%d", ch + 1, &k); n = strlen(ch + 1); get_SA(); for (int i = 2; i <= k; i++) { while (head <= tail && height[q[tail]] >= height[i]) --tail; q[++tail] = i; } for (int i = k; i <= n; i++) { if (i - q[head] + 1 >= k) ++head; while (head <= tail && height[q[tail]] >= height[i]) --tail; q[++tail] = i; int t = (k == 1) ? n - sa[i] + 1 : height[q[head]]; int g = max(height[i + 1], height[i - k + 1]); if (g <= t) ++cnt[g + 1], --cnt[t + 1]; } int maxn = 1, pos = -1; for (int i = 1; i <= n; i++) { cnt[i] += cnt[i - 1]; if (cnt[i] >= maxn) maxn = cnt[i], pos = i; } cout << pos << endl; } }
