用后缀数组求重复出现至少k次的可重叠最长子串的长度, 当然是可以用hash搞的,用后缀数组的话,只要在分组之后看看个数是不是大于等于k
#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack> #include <map> #include <set> #include <climits> #include <string> using namespace std; #define MP make_pair #define PB push_back typedef long long LL; typedef unsigned long long ULL; typedef vector<int> VI; typedef pair<int, int> PII; typedef pair<double, double> PDD; const int INF = INT_MAX / 3; const double eps = 1e-8; const LL LINF = 1e17; const double DINF = 1e60; const int maxn = 1e6 + 10; //以下是倍增法求后缀数组 int wa[maxn], wb[maxn], wv[maxn], ws[maxn]; int cmp(int *r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; } void da(int *r, int *sa, int n, int m) { int i, j, p, *x = wa, *y = wb, *t; for(i = 0; i < m; i++) ws[i] = 0; for(i = 0; i < n; i++) ws[x[i] = r[i]]++; for(i = 1; i < m; i++) ws[i] += ws[i - 1]; for(i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i; for(j = 1, p = 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++) ws[i] = 0; for(i = 0; i < n; i++) ws[wv[i]]++; for(i = 0; i < m; i++) ws[i] += ws[i - 1]; for(i = n - 1; i >= 0; i--) sa[--ws[wv[i]]] = y[i]; for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++; } } //以下是求解height 数组 int height[maxn], Rank[maxn]; void calheight(int *r, int *sa, int n) { int i, j, k = 0; for(i = 1; i <= n; i++) Rank[sa[i]] = i; for(i = 0; i < n; height[Rank[i++]] = k) for(k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++) ; } int n, k, val[maxn], sa[maxn]; bool ok(int len) { int nowcnt = 1; for(int i = 1; i <= n; i++) { if(height[i] >= len) { nowcnt++; if(nowcnt >= k) return true; } else nowcnt = 1; } return false; } int main() { while(scanf("%d%d", &n, &k) != EOF) { for(int i = 0; i < n; i++) scanf("%d", &val[i]), val[i]++; val[n] = 0; da(val, sa, n + 1, 1e6 + 5); calheight(val, sa, n); int l = 1, r = n, ans = 0; while(l <= r) { int mid = (l + r) >> 1; if(ok(mid)) { ans = mid; l = mid + 1; } else r = mid - 1; } printf("%d ", ans); } return 0; }