[BZOJ 1717][POJ3261]Milk Patterns[SA+二分]

还是2009那篇论文里介绍的做法

这里面有下载链接

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int k, n, rnk[20005], sa[20005], s[20005], H[20005], c[20005], K;

void Sort(int *x, int *y, int *rk) {
  static int C[20005];
  for (int i = 0; i <= k; ++i) C[i] = 0;
  for (int i = 1; i <= n; ++i) ++C[rk[i]];
  for (int i = 1; i <= k; ++i) C[i] += C[i - 1];
  for (int i = n; i; --i) y[C[rk[x[i]]]--] = x[i];
}

inline bool cmp(int *y, int a, int b, int m) {return y[a] == y[b] && y[a + m] == y[b + m];}
void get_SA() {
  static int Y[20005];
  int *y = Y, *rk = rnk;
  for (int i = 1; i <= n; ++i) y[i] = s[i];
  sort(y + 1, y + 1 + n);
  int m = unique(y + 1, y + 1 + n) - y - 1;
  for (int i = 1; i <= n; ++i) s[i] = lower_bound(y + 1, y + 1 + m, s[i]) - y;
  for (int i = 1; i <= n; ++i) rk[i] = s[y[i] = i];
  k = 20003;
  Sort(y, sa, rk);
  for (int m = 1, p = 0; p < n; k = p, m <<= 1) {
    for (p = 0; p < m; ++p) y[p + 1] = n - m + p + 1;
    for (int i = 1; i <= n; ++i)
      if (sa[i] > m) y[++p] = sa[i] - m;
    Sort(y, sa, rk), swap(y, rk);
    rk[sa[p = 1]] = 1;
    for (int i = 2; i <= n; ++i) rk[sa[i]] = cmp(y, sa[i - 1], sa[i], m) ? p : ++p;
  }
  for (int i = 1; i <= n; ++i) rnk[sa[i]] = i;
}

void get_H() {
  for (int i = 1, k = 0; i <= n; H[rnk[i++]] = k)
    for (k ? --k : 0; s[i + k] == s[sa[rnk[i] - 1] + k]; ++k);
}

bool check(int x) {
  for (int i = 2, cnt = 1; i <= n; ++i) {
    if (H[i] >= x) ++cnt;
    else cnt = 1;
    if (cnt == K) return 1;
  }
  return 0;
}

int main() {
  while (~scanf("%d%d", &n, &K) && n) {
    for (int i = 1; i <= n; ++i) scanf("%d", s + i), ++s[i];
    get_SA();
    get_H();
    int l = 1, r = n, mid;
    while (l <= r) {
      if (check(mid = l + r >> 1)) l = mid + 1;
      else r = mid - 1;
    }
    cout << l - 1 << endl;
  }
  return 0;
}