<p>You are given an <code>m x n</code> grid <code>grid</code> where:</p>

<ul>
	<li><code>&#39;.&#39;</code> is an empty cell.</li>
	<li><code>&#39;#&#39;</code> is a wall.</li>
	<li><code>&#39;@&#39;</code> is the starting point.</li>
	<li>Lowercase letters represent keys.</li>
	<li>Uppercase letters represent locks.</li>
</ul>

<p>You start at the starting point and one move consists of walking one space in one of the four cardinal directions. You cannot walk outside the grid, or walk into a wall.</p>

<p>If you walk over a key, you can pick it up and you cannot walk over a lock unless you have its corresponding key.</p>

<p>For some <code><font face="monospace">1 &lt;= k &lt;= 6</font></code>, there is exactly one lowercase and one uppercase letter of the first <code>k</code> letters of the English alphabet in the grid. This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.</p>

<p>Return <em>the lowest number of moves to acquire all keys</em>. If it is impossible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-keys2.jpg" style="width: 404px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@.a..&quot;,&quot;###.#&quot;,&quot;b.A.B&quot;]
<strong>Output:</strong> 8
<strong>Explanation:</strong> Note that the goal is to obtain all the keys not to open all the locks.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-key2.jpg" style="width: 404px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@..aA&quot;,&quot;..B#.&quot;,&quot;....b&quot;]
<strong>Output:</strong> 6
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-keys3.jpg" style="width: 244px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@Aa&quot;]
<strong>Output:</strong> -1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either an English letter, <code>&#39;.&#39;</code>, <code>&#39;#&#39;</code>, or <code>&#39;@&#39;</code>.&nbsp;</li>
	<li>There is exactly one&nbsp;<code>&#39;@&#39;</code>&nbsp;in the grid.</li>
	<li>The number of keys in the grid is in the range <code>[1, 6]</code>.</li>
	<li>Each key in the grid is <strong>unique</strong>.</li>
	<li>Each key in the grid has a matching lock.</li>
</ul>

#include <vector>
#include <queue>
#include <tuple>
#include <unordered_set>
#include <cmath>
#include <iostream>

class Solution {
public:
    int shortestPathAllKeys(std::vector<std::string> &grid) {
        int keyCnt = 0;
        std::pair<int, int> source;
        std::vector<std::pair<int, int>> keyCoors(6); // 假設最多有6把鑰匙
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (std::islower(grid[i][j])) {
                    keyCoors[grid[i][j] - 'a'] = {i, j};
                    ++keyCnt;
                } else if (grid[i][j] == '@') {
                    source = {i, j};
                }
            }
        }
        keyCoors.resize(keyCnt); // 去除未使用的鑰匙座標
        return aStarAlgorithm(source, keyCoors, grid);
    }

private:
    int aStarAlgorithm(const std::pair<int, int> &source, const std::vector<std::pair<int, int>> &keyCoors, const std::vector<std::string> &grid) {
        const std::vector<std::pair<int, int>> DIRS{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
        const int KEY_MASK = (1 << keyCoors.size()) - 1; // 所有鑰匙收集完畢的掩碼

        auto isInBoundary = [&grid](const int &x, const int &y) -> bool {
            return x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size();
        };

        // 啟發函數: 從當前位置到所有未收集鑰匙的曼哈頓距離
        auto hScore = [&keyCoors](const std::pair<int, int> &coor, const int &keys) -> int {
            int result = 0;
            for (int i = 0; i < keyCoors.size(); ++i) {
                if ((keys & (1 << i)) == 0) { // 只考慮未收集的鑰匙
                    result = std::max(result, std::abs(coor.first - keyCoors[i].first) + std::abs(coor.second - keyCoors[i].second));
                }
            }
            return result;
        };

        using NodeInfo = std::tuple<int, int, std::pair<int, int>, int>; // (fScore, gScore, (x, y), keys)
        struct customComp {
            bool operator()(const NodeInfo &a, const NodeInfo &b) const {
                return std::get<0>(a) > std::get<0>(b); // 比較 fScore
            }
        };

        std::priority_queue<NodeInfo, std::vector<NodeInfo>, customComp> openList;
        std::unordered_set<int> closedList;

        // 狀態哈希函數: 把 (x, y, keys) 壓縮為一個唯一的整數
        auto hashing = [](const std::pair<int, int> &coor, const int &keys) -> int {
            return coor.first << 12 | coor.second << 6 | keys;
        };

        openList.emplace(0 + hScore(source, 0), 0, source, 0); // 初始化，起點無鑰匙

        while (!openList.empty()) {
            int gScore = std::get<1>(openList.top());
            auto coor = std::get<2>(openList.top());
            int keys = std::get<3>(openList.top());
            openList.pop();

            // 如果所有鑰匙都已經收集完畢，返回步數
            if (keys == KEY_MASK) {
                return gScore;
            }

            int hashValue = hashing(coor, keys);
            // if (closedList.find(hashValue) != closedList.end()) {
            //     continue; // 避免重複擴展相同的狀態
            // }
            closedList.insert(hashValue);

            // 擴展當前節點的鄰居
            for (const auto &dir : DIRS) {
                int nx = coor.first + dir.first;
                int ny = coor.second + dir.second;

                if (isInBoundary(nx, ny) && grid[nx][ny] != '#') { // 確保不超出邊界且不是牆
                    if (std::isupper(grid[nx][ny]) && (keys & (1 << (grid[nx][ny] - 'A'))) == 0) {
                        continue; // 如果遇到門且沒有相應鑰匙，跳過
                    }

                    int newKeys = keys;
                    if (std::islower(grid[nx][ny])) { // 如果遇到鑰匙，更新狀態
                        newKeys |= 1 << (grid[nx][ny] - 'a');
                    }

                    int newHashValue = hashing({nx, ny}, newKeys);
                    if (closedList.find(newHashValue) == closedList.end()) {
                        int newGScore = gScore + 1;
                        int newFScore = newGScore + hScore({nx, ny}, newKeys);
                        openList.emplace(newFScore, newGScore, std::make_pair(nx, ny), newKeys);
                    }
                }
            }
        }

        return -1; // 如果無法收集所有鑰匙，返回 -1
    }
};

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