problem1 link

首先計算任意兩點的距離。然後列舉選出的集合中的兩個點，判斷其他點是否可以即可。

problem2 link

假設字串為$s$，長度為$n$。那麼對於$SA$中的兩個排名$SA_{i},SA_{i+1}$來說，應該儘量使得$s[SA_{i}]=s[SA_{i+1}]$。如果這個滿足的話，那麼需要兩個字尾滿足$s[SA_{i}+1\sim n-1]<s[SA_{i+1}+1\sim n-1]$,設他們的排名分別為$SA_{r},SA_{k}$，也就是說$r<k$即可。

problem3 link

code for problem1

#include <algorithm>
#include <unordered_set>
#include <vector>

class Egalitarianism3 {
 public:
  int maxCities(int n, const std::vector<int> &a, const std::vector<int> &b,
                const std::vector<int> &len) {
    if (n == 1) {
      return 1;
    }
    std::vector<std::vector<int>> g(n, std::vector<int>(n, -1));
    for (int i = 0; i < n - 1; ++i) {
      g[a[i] - 1][b[i] - 1] = g[b[i] - 1][a[i] - 1] = len[i];
    }
    for (int i = 0; i < n; ++i) {
      g[i][i] = 0;
    }
    for (int i = 0; i < n; ++i) {
      for (int u = 0; u < n; ++u) {
        if (g[u][i] != -1) {
          for (int v = 0; v < n; ++v) {
            if (g[i][v] != -1) {
              if (g[u][v] == -1 || g[u][v] > g[u][i] + g[i][v]) {
                g[u][v] = g[u][i] + g[i][v];
              }
            }
          }
        }
      }
    }
    auto Compute = [&](int s, int t) {
      std::unordered_set<int> all;
      all.insert(s);
      all.insert(t);
      const int d = g[s][t];
      for (int i = 0; i < n; ++i) {
        if (all.find(i) == all.end()) {
          bool tag = true;
          for (auto e : all) {
            if (g[i][e] != d) {
              tag = false;
              break;
            }
          }
          if (tag) {
            all.insert(i);
          }
        }
      }
      return static_cast<int>(all.size());
    };
    int result = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        result = std::max(result, Compute(i, j));
      }
    }
    return result;
  }
};

code for problem2

#include <vector>

class SuffixArrayDiv1 {
 public:
  int minimalCharacters(const std::vector<int> &SA) {
    int n = static_cast<int>(SA.size());
    std::vector<int> s(n + 1, -1);
    for (int i = 0; i < n; ++i) {
      s[SA[i]] = i;
    }
    int result = 1;
    for (int i = 0; i + 1 < n; ++i) {
      if (s[SA[i] + 1] > s[SA[i + 1] + 1]) {
        ++result;
      }
    }
    return result;
  }
};

code for problem3