AtCoder Regular Contest 050 D: Suffix Concat

問題

D: Suffix Concat - AtCoder Regular Contest 050 | AtCoder

解法

公式解説の通り。suffix array ではなく rolling hash でやった。
http://arc050.contest.atcoder.jp/data/arc/050/review.pdf

コード

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

class RollingHash {
 private:
  vector<long long> mod;
  void init() {
    // mod.push_back(999999937LL);
    mod.push_back(1000000009LL);
  }
  const long long base = 1000000007LL;
  int n;
  vector<vector<long long>> hs, pw;
  RollingHash() {}

 public:
  int cnt = 0;

  RollingHash(const string &s) : n(s.size()) {
    init();
    hs.resize(mod.size()), pw.resize(mod.size());
    for (int i = 0; i < mod.size(); i++) {
      hs[i].assign(n + 1, 0);
      pw[i].assign(n + 1, 0);
      hs[i][0] = 0;
      pw[i][0] = 1;
      for (int j = 0; j < n; j++) {
        // pw[i][j + 1] = pw[i][j] * base % mod[i];
        // hs[i][j + 1] = (hs[i][j] * base + s[j]) % mod[i];
        pw[i][j + 1] = pw[i][j] * base;
        hs[i][j + 1] = (hs[i][j] * base + s[j]);
      }
    }
  }

  inline long long hash(int l, int r, int i) {
    // return ((hs[i][r] - hs[i][l] * pw[i][r - l]) % mod[i] + mod[i]) % mod[i];
    return hs[i][r] - hs[i][l] * pw[i][r - l];
  }

  inline bool match(int l1, int r1, int l2, int r2) {
    bool ret = 1;
    for (int i = 0; i < mod.size(); i++)
      ret &= hash(l1, r1, i) == hash(l2, r2, i);
    return ret;
  }

  inline bool match(int l1, int l2, int k) {
    return match(l1, l1 + k, l2, l2 + k);
  }

  int lcp(int i, int j) {
    int l = 0, r = min(n - i, n - j) + 1;
    while (l + 1 < r) {
      int m = (l + r) / 2;
      if (match(i, i + m, j, j + m))
        l = m;
      else
        r = m;
    }
    return l;
  }
};

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N;
  string S;
  cin >> N >> S;
  RollingHash rh(S);

  vector<int> ans(N);
  iota(ans.begin(), ans.end(), 0);
  sort(ans.begin(), ans.end(), [&](const int &a, const int &b) {
    int i = a, j = b;
    if (i > j) swap(i, j);
    int lcp = rh.lcp(i, j);
    if (lcp < N - j) {
      if (S[i + lcp] < S[j + lcp])
        return (a == i);
      else
        return (a != i);
    } else {
      int nlcp = rh.lcp(i, N - (j - i));
      if (nlcp < j - i) {
        if (S[N - (j - i) + nlcp] < S[i + nlcp]) {
          return (a == i);
        } else
          return (a != i);
      } else {
        return a < b;
      }
    }
  });
  for (const int &a : ans) {
    cout << (a + 1) << "\n";
  }
}