首页 > 代码库 > SPOJ 687 Repeats 后缀数组
SPOJ 687 Repeats 后缀数组
和上一题差不多的方法。。没什么好说的
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = (5e4 + 10) * 4;#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x) - tb) * 3 + 2)int wa[maxn], wb[maxn], wv[maxn], ws[maxn];int c0(int *r, int a, int b) { return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];}int c12(int k, int *r, int a, int b) { if (k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1); else return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];}void sort(int *r, int *a, int *b, int n, int m) { int i; for (i = 0; i < n; i++) wv[i] = r[a[i]]; for (i = 0; i < m; i++) ws[i] = 0; for (i = 0; i < n; i++) ws[wv[i]]++; for (i = 1; i < m; i++) ws[i] += ws[i - 1]; for (i = n - 1; i >= 0; i--) b[--ws[wv[i]]] = a[i];}void dc3(int *r, int *sa, int n, int m) { int i, j, *rn = r + n, *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p; r[n] = r[n + 1] = 0; for (i = 0; i < n; i++) if (i % 3 != 0) wa[tbc++] = i; sort(r + 2, wa, wb, tbc, m); sort(r + 1, wb, wa, tbc, m); sort(r, wa, wb, tbc, m); for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++) rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++; if (p < tbc) dc3(rn, san, tbc, p); else for (i = 0; i < tbc; i++) san[rn[i]] = i; for (i = 0; i < tbc; i++) if (san[i] < tb) wb[ta++] = san[i] * 3; if (n % 3 == 1) wb[ta++] = n - 1; sort(r, wb, wa, ta, m); for (i = 0; i < tbc; i++) wv[wb[i] = G(san[i])] = i; for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++) sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++]; for (; i < ta; p++) sa[p] = wa[i++]; for (; j < tbc; p++) sa[p] = wb[j++];}int Rank[maxn], height[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++);}#undef F#undef Gint str[maxn], sa[maxn], T, n, minv[maxn][30];void init_RMQ() { for(int i = 0; i <= n; i++) { minv[i][0] = height[i]; } for(int j = 1; (1 << j) <= n + 1; j++) { for(int i = 0; i + (1 << j) - 1 <= n; i++) { minv[i][j] = min(minv[i][j - 1], minv[i + (1 << (j - 1))][j - 1]); } }}int query(int ql, int qr) { if(ql > qr) swap(ql, qr); ql++; int k = 0; while((1 << (k + 1)) <= qr - ql + 1) k++; return min(minv[ql][k], minv[qr - (1 << k) + 1][k]);}int main() { scanf("%d", &T); while(T--) { scanf("%d", &n); for(int i = 0; i < n; i++) { char tmp; scanf(" %c", &tmp); str[i] = tmp; } str[n] = 0; dc3(str, sa, n + 1, 200); calheight(str, sa, n); init_RMQ(); int maxstep = 1; for(int L = 1; L <= n; L++) { for(int i = 0; i + L < n; i += L) { int s1 = query(Rank[i], Rank[i + L]); int step = s1 / L + 1, ql = i - (L - s1 % L); if(ql >= 0) { step = max(step, query(Rank[ql], Rank[ql + L]) / L + 1); } maxstep = max(maxstep, step); } } printf("%d\n", maxstep); } return 0;}
SPOJ 687 Repeats 后缀数组
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。