C++ solution with Bitmask DP and BFS precomputation.
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(((m * n) ^ 2) + ((p * p) * (2 ^ p)))\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(((m * n) ^ 2) + ((p * p) * (2 ^ p)))\np -> positions.size(), m == n == 50 (grid dimensions)\n\n# Code\n```cpp []\nint mem[51][51][51][51];\nbool flag = false;\nint dp[16][(int)(1 << 16)];\nconst int N = 50;\nconst int INF = 1e9;\nint dx[] = {2, 2, -2, -2, 1, 1, -1, -1};\nint dy[] = {1, -1, 1, -1, 2, -2, 2, -2};\n\nclass Solution {\nprivate:\n    int fullMask, n;\npublic:\n    void bfs(int sx, int sy) {\n        int dist[N][N];\n        for(int i = 0; i < N; i++)\n            for(int j = 0; j < N; j++)\n                dist[i][j] = INF;\n        \n        queue<pair<int, int>> q;\n        q.push({sx, sy});\n        dist[sx][sy] = 0;\n        int level = 0;\n        \n        while (!q.empty()) {\n            int sz = q.size();\n            while(sz--){\n                int x = q.front().first;\n                int y = q.front().second;\n                q.pop();\n                \n                for (int i = 0; i < 8; ++i) {\n                    int nx = x + dx[i];\n                    int ny = y + dy[i];\n                    \n                    if (nx >= 0 && nx < N && ny >= 0 && ny < N && dist[nx][ny] == INF) {\n                        dist[nx][ny] = 1 + level;\n                        q.push({nx, ny});\n                    }\n                }\n            }\n            level++;\n        }\n        \n        for (int x = 0; x < N; ++x) {\n            for (int y = 0; y < N; ++y) {\n                mem[sx][sy][x][y] = dist[x][y];\n            }\n        }\n    }\n    void init(){\n        if(flag) return;\n        flag = true;\n\n        memset(mem, -1, sizeof(mem));\n        for (int x1 = 0; x1 < N; ++x1) {\n            for (int y1 = 0; y1 < N; ++y1) {\n                bfs(x1, y1);\n            }\n        }\n    }\n    int movesMade(int x1, int y1, int x2, int y2){\n        return mem[x1][y1][x2][y2];\n    }\n    int solve(int prev, int mask, bool turn, vector<vector<int>> &positions){\n        if(mask == fullMask) return 0;\n\n        if(dp[prev][mask] != -1)\n            return dp[prev][mask];\n\n        int ans = turn ? INF : -INF;\n        for(int i = 0; i < positions.size(); i++){\n            if(mask & (1 << i)) continue;\n            int px = positions[prev][0];\n            int py = positions[prev][1];\n            int nx = positions[i][0];\n            int ny = positions[i][1];\n            if(turn)\n                ans = min(movesMade(px, py, nx, ny) + solve(i, mask | (1 << i), !turn, positions), ans);\n            else   \n                ans = max(movesMade(px, py, nx, ny) + solve(i, mask | (1 << i), !turn, positions), ans);\n        }\n        return dp[prev][mask] = ans;\n    }\n    int maxMoves(int kx, int ky, vector<vector<int>>& positions) {\n        init();\n        n = positions.size();\n        fullMask = (1 << n) - 1;\n        for(int prev = 0; prev < n; prev++)\n            for(int mask = 0; mask <= fullMask; mask++)\n                dp[prev][mask] = -1;\n\n        int ans = -INF;\n        for(int i = 0; i < positions.size(); i++){\n            auto curr = positions[i];\n            ans = max(movesMade(kx, ky, curr[0], curr[1]) + solve(i, 1 << i, 1, positions), ans);\n        }\n        return ans;\n    }\n};\n```
['C++']