problem1 link
选择所有的'+'或者所有的‘-’,一定是这两种中的一种最大。
problem2 link
首先,第$n$个盘子初始时所在的柱子一定的最后所有的盘子都应该挪到的柱子。所以,可以枚举第$n$个盘子在哪个柱子上。
假设目前枚举第$n$个盘子在第三个柱子上,那么假设要求解的问题为 $F(k,x,y,z,3)$, 其中$x=count[0],y=count[1],z=count[2]-1$,表示处理剩下的$n-1=x+y+z$个盘子,三根柱子上的个数为$(x,y,z)$。
那么对于第$n-1$个盘子,如果也放到第三根柱子上,那么接下来就是求解$F(k,x,y,z-1,3)$。如果将其放置到第二根柱子上,那么要首先把把前面的$n-2$个盘子挪到第一根柱子上,然后挪$n-1$盘子,然后再把$n-2$个盘子挪到第三根上,这三步分别需要:$F(k^{'},x,y-1,z,1),1,2^{n-2}-1$ ,所以$k^{'}=k-2^{n-2}=k-2^{x+y+z-1}$,那么接下来要求解的是$F(k-2^{x+y+z-1},x,y-1,z,1)$。
按照上面的思路可以得到结论如果$kgeq 2^{x+y+z-1}$,那么此时的盘子一定要换一根柱子。
problem3 link
这个就是直接搜索就可以了。把原来的树构造出来。
code for problem1
#include <algorithm> #include <string> class MaximumRange { public: int findMax(const std::string &s) { int n = static_cast<int>(s.size()); int x = 0; for (auto c : s) { if (c == '+') { ++x; } } return std::max(x, n - x); } };
code for problem2
#include <string> #include <vector> class ClassicTowers { public: std::string findTowers(long long k, const std::vector<int> &count) { if (Check(k, count[0], count[1], count[2], "ABC")) { return result; } if (Check(k, count[0], count[2], count[1], "ACB")) { return result; } if (Check(k, count[1], count[2], count[0], "BCA")) { return result; } return ""; } private: std::string result; std::vector<int> all; bool Dfs(long long k, int x, int y, int z, int idx) { if (x + y + z == 0) { return k == 0; } int curr = x + y + z - 1; if ((k & (1ll << curr)) == 0) { all[curr] = idx; if (idx == 1 && x > 0) { return Dfs(k, x - 1, y, z, idx); } else if (idx == 2 && y > 0) { return Dfs(k, x, y - 1, z, idx); } else if (idx == 3 && z > 0) { return Dfs(k, x, y, z - 1, idx); } else { return false; } } else { for (int i = 1; i <= 3; ++i) { if (i != idx) { all[curr] = i; if (i == 1 && x > 0 && Dfs(k ^ (1ll << curr), x - 1, y, z, 6 - idx - i)) { return true; } if (i == 2 && y > 0 && Dfs(k ^ (1ll << curr), x, y - 1, z, 6 - idx - i)) { return true; } if (i == 3 && z > 0 && Dfs(k ^ (1ll << curr), x, y, z - 1, 6 - idx - i)) { return true; } } } return false; } } bool Check(long long k, int x, int y, int z, const std::string &table) { if (z == 0) { return false; } long long maxk = (1ll << (x + y + z - 1)) - 1; if (k > maxk) { return false; } all.resize(x + y + z); all.back() = 3; if (Dfs(k, x, y, z - 1, 3)) { result = ""; for (int i = 0; i < x + y + z; ++i) { result += table[all[i] - 1]; } return true; } return false; } };
code for problem3
#include <algorithm> #include <map> #include <memory> #include <string> #include <vector> class PreInPost { public: std::vector<int> findMissing(const std::vector<std::string> &s, const std::vector<int> &a1, const std::vector<int> &a2, const std::string &e1, const std::string &e2) { table_s = s; auto status = Dfs(a1, a2, e1, e2); if (!status.flag) { return {}; } std::vector<int> result; std::string mode = "pre"; for (int i = 0; i < 6; i += 2) { if (s[i] != e1 && s[i] != e2) { Order(status.root.get(), s[i], &result); break; } } return result; } private: struct Node { int idx = -1; std::shared_ptr<Node> left = nullptr; std::shared_ptr<Node> right = nullptr; void SetRoot(int x) { if (idx == -1) { idx = x; } else if (idx != x) { idx = -2; } } }; struct Status { std::shared_ptr<Node> root = nullptr; bool flag = false; }; struct KeyNode { std::vector<int> a1; std::vector<int> a2; std::string e1; std::string e2; KeyNode() = default; KeyNode(const std::vector<int> &a1, const std::vector<int> &a2, const std::string &e1, const std::string &e2) : a1(a1), a2(a2), e1(e1), e2(e2) {} bool operator<(const KeyNode &key) const { if (a1 != key.a1) { return a1 < key.a1; } if (a2 != key.a2) { return a2 < key.a2; } if (e1 != key.e1) { return e1 < key.e1; } return e2 < key.e2; } }; std::map<KeyNode, Status> visited_states; std::vector<std::string> table_s; int LeftModeIdx(const std::string &e) { if (e == "pre") { return 0; } else if (e == "in") { return 2; } else { return 4; } } void Order(const Node *node, const std::string &e, std::vector<int> *result) { if (node == nullptr) { return; } if (e == "pre") { result->push_back(node->idx); Order(node->left.get(), table_s[0], result); Order(node->right.get(), table_s[1], result); } else if (e == "in") { Order(node->left.get(), table_s[2], result); result->push_back(node->idx); Order(node->right.get(), table_s[3], result); } else { Order(node->left.get(), table_s[4], result); Order(node->right.get(), table_s[5], result); result->push_back(node->idx); } } bool SameSet(const std::vector<int> &a1, const std::vector<int> &a2) { long long s[4] = {0, 0, 0, 0}; constexpr int kEach = 60; for (size_t i = 0; i < a1.size(); ++i) { s[a1[i] / kEach] ^= 1ll << (a1[i] % kEach); s[a2[i] / kEach] ^= 1ll << (a2[i] % kEach); } return s[0] == 0 && s[1] == 0 && s[2] == 0 && s[3] == 0; } Status Dfs(const std::vector<int> &a1, const std::vector<int> &a2, const std::string &e1, const std::string &e2) { Status status; if (a1.empty()) { status.flag = true; return status; } status.root = std::shared_ptr<Node>(new Node); auto Set = [&](const std::vector<int> &a, const std::string &e) { if (e == "pre" || e == "post") { status.root->SetRoot(e == "pre" ? a.front() : a.back()); } }; Set(a1, e1); Set(a2, e2); if (status.root->idx == -2) { return status; } KeyNode key_node(a1, a2, e1, e2); if (visited_states.find(key_node) != visited_states.end()) { return visited_states[key_node]; } std::vector<int> new_a1 = a1; std::vector<int> new_a2 = a2; int m = -1; auto RemoveRoot = [&](const std::string &e, std::vector<int> *a) { if (e == "pre") { a->erase(a->begin()); } else if (e == "post") { a->pop_back(); } else { m = static_cast<int>(std::find(a->begin(), a->end(), status.root->idx) - a->begin()); a->erase(a->begin() + m); } }; RemoveRoot(e1, &new_a1); RemoveRoot(e2, &new_a2); if (!SameSet(new_a1, new_a2)) { return visited_states[key_node] = status; } int n = static_cast<int>(new_a1.size()); std::vector<int> right1; std::vector<int> right2; auto Check = [&]() { if (SameSet(new_a1, new_a2) && SameSet(right1, right2)) { Status left = Dfs(new_a1, new_a2, table_s[LeftModeIdx(e1)], table_s[LeftModeIdx(e2)]); Status right = Dfs(right1, right2, table_s[LeftModeIdx(e1) + 1], table_s[LeftModeIdx(e2) + 1]); if (left.flag && right.flag) { status.root->left = left.root; status.root->right = right.root; status.flag = true; return true; } } return false; }; if (m == -1) { if (!Check()) { for (int i = n - 1; i >= 0; --i) { right1.insert(right1.begin(), new_a1.back()); right2.insert(right2.begin(), new_a2.back()); new_a1.pop_back(); new_a2.pop_back(); if (Check()) { break; } } } } else { for (int i = m; i < n; ++i) { right1.push_back(new_a1[i]); right2.push_back(new_a2[i]); } new_a1.erase(new_a1.begin() + m, new_a1.end()); new_a2.erase(new_a2.begin() + m, new_a2.end()); Check(); } return visited_states[key_node] = status; } };