[深度优先搜索] leetcode 864 Shortest Path to Get All Keys

problem:https://leetcode.com/problems/shortest-path-to-get-all-keys/

       这道题其它地方都挺基础的，就是普通的宽搜，难点主要在于visit状态的维护。

       如果不考虑visit数组，则会出现大量重复的来回搜索，比如刚刚从1->2，下一步又从2->1。

       但是，已经访问过的地方，有可能再次被访问，比如先取得了钥匙a，因为前面没有路了，只能原路返回，再去取钥匙b。所以这意味着我们不能单独以坐标作为visit记录，而是需要考虑当前钥匙状态——没有拿到钥匙a和拿着钥匙a原路返回，这两者是存在差异的。前者是我们不期望的，后者则是我们需要重复搜索的（为了找到答案）。

class Solution {
public:
    int m, n;
    vector<int> dx{ 0,1,0,-1 };
    vector<int> dy{ 1,0,-1,0 };

    bool checkKey(unsigned int key, char lock)
    {
        unsigned int data = lock - 'A';
        return (key & (1 << data)) != 0;
    }

    int shortestPathAllKeys(vector<string>& grid) {
        m = grid.size();
        n = grid[0].size();
        
        queue<pair<int, unsigned int>> q;
        unsigned int target = 0;
        
        vector<vector<int>> visit(m * n, vector<int>(200,false));
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (grid[i][j] == '@')
                {
                    visit[i * n + j][0] = true;
                    q.push({ i * n + j, 0 });
                }
                else if (grid[i][j] >= 'a' && grid[i][j] <= 'f')
                {
                    target |= (1 << (grid[i][j] - 'a'));
                }
            }
        }
        
        int res = 0;
        while (!q.empty())
        {
            int size = q.size();
            res++;
            while (size--)
            {
                auto cur = q.front();
                q.pop();
                int i = cur.first / n;
                int j = cur.first % n;
                unsigned int key = cur.second;

                for (int k = 0; k < 4; k++)
                {
                    int x = i + dx[k];
                    int y = j + dy[k];
                    if (x >= 0 && y >= 0 && x < m && y < n)
                    {
                        if(grid[x][y] == '#') continue;
                        int next = x * n + y;
                        int newKey = key;
                        if (grid[x][y] >= 'A' && grid[x][y] <= 'F')
                        {
                            if (!checkKey(key, grid[x][y]))
                            {
                                continue;
                            }
                        }
                        else if (grid[x][y] >= 'a' && grid[x][y] <= 'f')
                        {
                            newKey = key | (1 << (grid[x][y] - 'a'));
                            if (newKey == target)
                            {
                                return res;
                            }
                        }
                        
                        if(!visit[next][newKey])
                        {
                            visit[next][newKey] = true;
                            q.push({next, newKey});
                        }
                    }
                }
            }

        }

        return -1;
    }
};

标签：Get,Keys,visit,864,int,vector,grid,&&,钥匙
来源： https://www.cnblogs.com/fish1996/p/11272647.html