poj 3693(后缀数组)
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题意:一个由小写字母组成的字符串,问这个字符串内重复次数最多的连续子串,多组解输出字典序最小的。
题解:这里算法合集之《后缀数组——处理字符串的有力工具》例8讲的很明白,先枚举重复子串的长度为L,以s[0],s[L],s[2 * L]…为起点,有两个相邻的长度为L的串重复出现,那么就可以用height求两个后缀的lcp,lcp / L + 1是次数,但不一定是最大次数,还要从首字符向前匹配到不能匹配为止,一旦向前能再匹配出L - lcp % L个字符,说明次数还可以加1。如果重复次数相同用rank判断字典序。L - lcp % L的思路出自这里http://blog.csdn.net/acm_cxlove/article/details/7941205
#include <cstdio>#include <cstring>#include <algorithm>#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x) - tb) * 3 + 2)using namespace std;const int N = 100005;int wa[N], wb[N], ws[N], wv[N], sa[N * 3];int rank[N * 3], height[N * 3], s[N], f[N * 3][35];char str[N];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); 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) { for (int i = 0; i < n; i++) wv[i] = r[a[i]]; for (int i = 0; i < m; i++) ws[i] = 0; for (int i = 0; i < n; i++) ws[wv[i]]++; for (int i = 1; i < m; i++) ws[i] += ws[i - 1]; for (int 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++];}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++);}void RMQ_init(int cnt) { for (int i = 0; i < cnt; i++) f[i][0] = height[i]; for (int j = 1; (1 << j) <= cnt; j++) for (int i = 0; i + (1 << j) - 1 < cnt; i++) f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);}int RMQ(int L, int R) { int k = 0; while (1 << (k + 1) <= R - L + 1) k++; return min(f[L][k], f[R - (1 << k) + 1][k]);}int main() { int cas = 1; while (scanf("%s", str) == 1 && str[0] != '#') { int len = 0; char c = 'z' + 1; for (len = 0; str[len]; len++) { s[len] = str[len] - 'a' + 1; c = min(c, str[len]); } s[len] = 0; dc3(s, sa, len + 1, 50); calheight(s, sa, len); RMQ_init(len + 1); int maxx = 0, st = 0, Len = 0; for (int i = 1; i <= len / 2; i++) { for (int j = 0; j + i < len; j += i) { int pos1 = rank[j], pos2 = rank[j + i]; if (pos1 > pos2) swap(pos1, pos2); int temp = RMQ(pos1 + 1, pos2); int num = temp / i + 1; int r = i - temp % i; int pos = j, cnt = 0; for (int k = j - 1; k > j - i && str[k] == str[k + i]; k--) { cnt++; if (cnt == r) num++, pos = k; else if (rank[pos] > rank[k]) pos = k; } if (num > maxx) { maxx = num; Len = num * i; st = pos; } else if (num == maxx && rank[st] > rank[pos]) { st = pos; Len = num * i; } } } if (!maxx) printf("Case %d: %c\n", cas++, c); else { printf("Case %d: ", cas++); for (int i = st; i < st + Len; i++) printf("%c", str[i]); printf("\n"); } } return 0;}
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