topcoder srm 580 div1

problem1 link

最优选择一定是在$2n$个端点中选出两个。

problem2 link

分开考虑每个区间。设所有区间的左端点的最大值为$lc$,所有区间的右端点的最小值为$rc$.对于某个区间$L$，其实就是找最少的区间（包括$L$）能够完全覆盖区间$[lc,rc]$.

设$L$的左右端点为$L_{1},L_{2}$。考虑从$L_{1}$向左扩展，每次扩展一定是找一个区间$p$满足$p_{2}geq L_{1}$且使得$p_{1}$最小。右端点向右扩展类似。

problem3 link

对于后手来说，它一定是每次只放置一个障碍格子在先手目前的位置。否则，如果放置了太多，那么先手就容易决策出向左还是向右比较好。

对于先手来说，假设目前先手在$(r,c)$，那么如果$(r,c)$和$(r+1,c)$之间没有障碍，且轮到先手，应该直接跳到$(r+1,c)$,否则，那么由后手的策略来看，此时已经有的障碍一定是一个包含$(r,c)$的区间，此时先手可以尝试向左或者向右运动到障碍区间外一个格子。

这样就可以用$(r,c,left,right,tag)$来表示一个状态来进行动态规划。$(r,c)$表示先手目前的位置，$(left, right)$表示目前障碍的区间，$tag$指示目前是先手还是后手。

code for problem1

#include <vector>

class EelAndRabbit {
 public:
  int getmax(std::vector<int> &l, std::vector<int> t) {
    const int n = static_cast<int>(l.size());
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 2; ++j) {
        int p1 = j == 0 ? -t[i] : -t[i] - l[i];
        for (int k = 0; k < n; ++k) {
          for (int r = 0; r < 2; ++r) {
            int p2 = r == 0 ? -t[k] : -t[k] - l[k];
            if (p1 != p2) {
              int num = 0;
              long long mask = 0;
              for (int x = 0; x < n; ++x) {
                if ((-t[x] - l[x] <= p1 && p1 <= -t[x]) ||
                    (-t[x] - l[x] <= p2 && p2 <= -t[x])) {
                  if ((mask & (1ll << x)) == 0) {
                    ++num;
                    mask |= 1ll << x;
                  }
                }
              }
              ans = std::max(ans, num);
            }
          }
        }
      }
    }
    return ans;
  }
};

code for problem2

#include <algorithm>
#include <string>
#include <vector>

constexpr int N = 2500;
constexpr int M = 10000;

int s[N], t[N];
int back[M], next[M];
int back_cost[M], next_cost[M];
int sort_idx[N];

class ShoutterDiv1 {
 public:
  int count(const std::vector<std::string>& s1000,
            const std::vector<std::string>& s100,
            const std::vector<std::string>& s10,
            const std::vector<std::string>& s1,
            const std::vector<std::string>& t1000,
            const std::vector<std::string>& t100,
            const std::vector<std::string>& t10,
            const std::vector<std::string>& t1) {
    int n = 0;
    for (const auto& e : s1000) {
      n += static_cast<int>(e.size());
    }
    for (int i = 0; i < n; ++i) {
      s[i] = t[i] = 0;
      sort_idx[i] = i;
    }
    for (int i = 0; i < M; ++i) {
      back[i] = next[i] = i;
    }
    Parse(s1000, s, 1000);
    Parse(s100, s, 100);
    Parse(s10, s, 10);
    Parse(s1, s, 1);
    Parse(t1000, t, 1000);
    Parse(t100, t, 100);
    Parse(t10, t, 10);
    Parse(t1, t, 1);

    int c_left = t[0];
    int c_right = s[0];
    for (int i = 0; i < n; ++i) {
      back[t[i]] = std::min(back[t[i]], s[i]);
      next[s[i]] = std::max(next[s[i]], t[i]);
      c_left = std::min(c_left, t[i]);
      c_right = std::max(c_right, s[i]);
    }
    for (int i = M - 2; i >= 0; --i) {
      back[i] = std::min(back[i], back[i + 1]);
    }
    for (int i = 1; i < M; ++i) {
      next[i] = std::max(next[i], next[i - 1]);
    }
    for (int i = 0; i < M; ++i) {
      if (i <= c_left) {
        back_cost[i] = 0;
      } else if (i == back[i]) {
        back_cost[i] = -1;
      } else {
        if (back_cost[back[i]] == -1) {
          back_cost[i] = -1;
        } else {
          back_cost[i] = 1 + back_cost[back[i]];
        }
      }
    }
    for (int i = M - 1; i >= 0; --i) {
      if (i >= c_right) {
        next_cost[i] = 0;
      } else if (i == next[i]) {
        next_cost[i] = -1;
      } else {
        if (next_cost[next[i]] == -1) {
          next_cost[i] = -1;
        } else {
          next_cost[i] = 1 + next_cost[next[i]];
        }
      }
    }
    auto Cost = [&](int left, int right) {
      if (back_cost[left] == -1 || next_cost[right] == -1) {
        return -1;
      }
      return back_cost[left] + next_cost[right];
    };

    std::sort(sort_idx, sort_idx + n, [&](int x, int y) {
      return s[x] < s[y] || (s[x] == s[y] && t[x] < t[y]);
    });

    int result = 0;
    for (int i = 0; i < n; ++i) {
      int tmp = Cost(s[sort_idx[i]], t[sort_idx[i]]);
      for (int j = 0; j < N && s[sort_idx[j]] <= s[sort_idx[i]]; ++j) {
        if (t[sort_idx[i]] <= t[sort_idx[j]]) {
          int val = Cost(s[sort_idx[j]], t[sort_idx[j]]);
          if (val != -1 && (tmp == -1 || tmp > 1 + val)) {
            tmp = 1 + val;
          }
        }
      }
      if (tmp == -1) {
        return -1;
      }
      result += tmp;
    }
    return result;
  }

 private:
  void Parse(const std::vector<std::string>& d, int* data, int base) {
    int idx = 0;
    for (size_t i = 0; i < d.size(); ++i) {
      for (size_t j = 0; j < d[i].size(); ++j) {
        data[idx++] += (d[i][j] - '0') * base;
      }
    }
  }
};

code for problem3

#include <cstring>
#include <limits>
#include <string>
#include <vector>

#include <iostream>

constexpr int N = 50;
constexpr int kEachUseBit = 16;
constexpr int kRabbitTagBit = 31;
constexpr int kEelTagBit = 15;
constexpr int kMask = (1 << (kEachUseBit - 1)) - 1;

using TType = unsigned int;

TType rabbit_eel[N][N][N][N + 1];
int cost[N][N];
int h, w;

class WallGameDiv1 {
 public:
  int play(const std::vector<std::string> &costs) {
    h = static_cast<int>(costs.size());
    w = static_cast<int>(costs[0].size());
    for (int i = 0; i < h; ++i) {
      cost[i][0] = costs[i][0] - '0';
      for (int j = 1; j < w; ++j) {
        cost[i][j] = cost[i][j - 1] + costs[i][j] - '0';
      }
    }

    memset(rabbit_eel, 0, sizeof(rabbit_eel));
    // Iterator each row to avoid more depths of recursion.
    for (int i = h - 1; i >= 0; --i) {
      for (int j = 0; j < w; ++j) {
        Eel(i, j, j, j);
      }
    }
    int result = std::numeric_limits<int>::max();
    for (int i = 0; i < w; ++i) {
      result = std::min(result, Cost(0, i, i) + Eel(0, i, i, i));
    }
    return result;
  }

 private:
  bool Computed(const TType &val, bool is_rabbit) {
    if (is_rabbit) {
      return (val & (1u << kRabbitTagBit)) != 0;
    } else {
      return (val & (1u << kEelTagBit)) != 0;
    }
  }

  int GetCost(const TType &val, bool is_rabbit) {
    if (!Computed(val, is_rabbit)) {
      return -1;
    }
    if (is_rabbit) {
      return static_cast<int>((val >> kEachUseBit) & kMask);
    } else {
      return static_cast<int>(val & kMask);
    }
  }

  int SetCost(TType &val, bool is_rabbit, int cost) {
    if (is_rabbit) {
      val |= (static_cast<TType>(cost)) << kEachUseBit;
      val |= 1u << kRabbitTagBit;
    } else {
      val |= cost;
      val |= 1u << kEelTagBit;
    }
    return cost;
  }

  int Eel(int row, int col, int left, int right) {
    TType &val = rabbit_eel[row][col][left][right];
    int result = GetCost(val, false);
    if (result != -1) {
      return result;
    }
    if (row == h - 1) {
      return SetCost(val, false, 0);
    }
    result = Rabbit(row, col, left, right);

    if (right - left < w - 1) {
      result = std::max(result, Rabbit(row, col, std::min(col, left),
                                       std::max(col + 1, right)));
    }

    return SetCost(val, false, result);
  }

  int Rabbit(int row, int col, int left, int right) {
    TType &val = rabbit_eel[row][col][left][right];
    int result = GetCost(val, true);
    if (result != -1) {
      return result;
    }
    if ((left <= col) && (col < right)) {
      result = std::numeric_limits<int>::max();
      if (left > 0) {
        result = Cost(row, left - 1, col - 1) + Eel(row, left - 1, left, right);
      }
      if (right < w) {
        result = std::min(
            result, Cost(row, col + 1, right) + Eel(row, right, left, right));
      }
    } else {
      result = Cost(row + 1, col, col) + Eel(row + 1, col, col, col);
    }
    return SetCost(val, true, result);
  }

  int Cost(int row, int left, int right) {
    if (left <= right) {
      if (left == 0) {
        return cost[row][right];
      }
      return cost[row][right] - cost[row][left - 1];
    }
    return 0;
  }
};