题意:训练指南P225
分析:二分寻找长度,用hash值来比较长度为L的字串是否相等。
#include <bits/stdc++.h> using namespace std; typedef unsigned long long ull; const int N = 4e4 + 5; const int x = 123; ull H[N], _hash[N], xp[N]; int rk[N]; char str[N]; int m; void get_hash(char *s, int len) { H[len] = 0; for (int i=len-1; i>=0; --i) { H[i] = H[i+1] * x + (s[i] - 'a'); } xp[0] = 1; for (int i=1; i<len; ++i) { xp[i] = xp[i-1] * x; } } bool cmp(const int &a, const int &b) { return (_hash[a] < _hash[b] || (_hash[a] == _hash[b] && a < b)); } int check(int L, int len) { int cnt = 0, pos = -1, c = 0; for (int i=0; i<len-L+1; ++i) { rk[i] = i; _hash[i] = H[i] - H[i+L] * xp[L]; } sort (rk, rk+len-L+1, cmp); for (int i=0; i<len-L+1; ++i) { if (i == 0 || _hash[rk[i]] != _hash[rk[i-1]]) c = 0; if (++c >= m) pos = max (pos, rk[i]); } return pos; } int main(void) { while (scanf ("%d", &m) == 1) { if (!m) break; scanf ("%s", &str); int len = strlen (str); get_hash (str, len); if (check (1, len) == -1) puts ("none"); else { int l = 1, r = len + 1; while (r - l > 1) { int mid = l + r >> 1; if (check (mid, len) >= 0) l = mid; else r = mid; } printf ("%d %d ", l, check (l, len)); } } return 0; }
后缀数组也可以求解,具体就是二分答案,height数组分组判断是否满足存在题意的解,并使最优。(m=1时特判处理)
#include <bits/stdc++.h> const int N = 4e4 + 5; int sa[N], rank[N], height[N]; int ws[N], wa[N], wb[N]; char s[N]; bool cmp(int *r, int a, int b, int l) { return (r[a] == r[b] && r[a+l] == r[b+l]); } void DA(char *r, int n, int m = 128) { int i, j, p, *x = wa, *y = wb; 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<m; ++i) ws[i] = 0; for (i=0; i<n; ++i) ws[x[y[i]]]++; for (i=1; i<m; ++i) ws[i] += ws[i-1]; for (i=n-1; i>=0; --i) sa[--ws[x[y[i]]]] = y[i]; std::swap (x, y); for (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++; } } } void calc_height(char *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; ++i) { if (k) k--; j = sa[rank[i]-1]; while (r[i+k] == r[j+k]) k++; height[rank[i]] = k; } } int m; int check(int len, int n) { int p = -1; int cnt = 0, ret = -1; for (int i=1; i<=n; ++i) { if (height[i] >= len) { if (p == -1) { p = std::max (sa[i-1], sa[i]); } else { p = std::max (p, std::max (sa[i-1], sa[i])); } cnt++; if (cnt + 1 >= m) { ret = std::max (ret, p); } } else { p = -1; cnt = 0; } } return ret; } int main() { while (scanf ("%d", &m) == 1) { if (!m) break; scanf ("%s", s); int n = strlen (s); if (m == 1) { printf ("%d %d ", n, 0); continue; } DA (s, n + 1); calc_height (s, sa, n); int best = 0, pos = -1; int left = 0, right = n; while (left <= right) { int mid = left + right >> 1; int res = check (mid, n); if (res != -1) { if (best < mid) { best = mid; pos = res; } else if (mid > 0 && best == mid && pos < res) { pos = res; } left = mid + 1; } else { right = mid - 1; } } if (pos == -1) { puts ("none"); } else { printf ("%d %d ", best, pos); } } return 0; }