C++ DP with bitmask and game strategy Question
Approach: Using BFS and Dynamic Programming with Bitmasking\n\nIn this problem, we need to calculate the minimum or maximum number of moves required for a knigh
### Approach: Using BFS and Dynamic Programming with Bitmasking\n\nIn this problem, we need to calculate the minimum or maximum number of moves required for a knight on a chessboard to visit all given positions. The solution utilizes Breadth-First Search (BFS) to compute minimum distances between positions and Dynamic Programming (DP) with Bitmasking to find the optimal solution.\n\n1. BFS for Minimum Distance Calculation\nWe first use BFS to compute the minimum number of moves required for the knight to travel from one position to another. Here\'s how we do it:\n\nDirections for Knight Moves: We define all possible moves that a knight can make on the chessboard.\nBFS Function: For each starting position (x, y), we compute the shortest distance to all other positions using BFS.\nKey Insight: BFS is ideal here because it explores all possible moves level by level, ensuring that we find the shortest path in an unweighted grid.\n\n2. Dynamic Programming with Bitmasking\nTo determine the minimum or maximum moves needed, we use DP with bitmasking:\n\n### State Definition:\n\ndp[idx][mask]: Minimum moves needed starting from position idx with the remaining positions represented by mask.\nAlice or Bob turns are used to determine whether we are maximizing or minimizing the moves.\nRecursive Function:\n\nWe recursively compute the result by transitioning through all unvisited positions and updating the mask accordingly.\nThe function returns the optimal number of moves based on whose turn it is.\n\n### CODE:\n```\nclass Solution {\npublic:\n    vector<vector<int>> directions = {{2, 1}, {2, -1}, {-2, 1}, {-2, -1}, {1, 2}, {1, -2}, {-1, 2}, {-1, -2}};\n\n    void BFS(int x, int y, int index, vector<vector<int>>& minDist, vector<vector<int>>& pos) {\n        vector<vector<int>> t(50, vector<int>(50, -1));\n        queue<pair<int, int>> que;\n        que.push({x, y});\n        t[x][y] = 0;\n\n        while(!que.empty()) {\n            auto[currX, currY] = que.front();\n            que.pop();\n\n            for(auto &dir : directions) {\n                int newX = currX + dir[0];\n                int newY = currY + dir[1];\n\n                if(newX >= 0 && newX < 50 && newY >= 0 && newY < 50 && t[newX][newY] == -1) {\n                    t[newX][newY] = t[currX][currY] + 1;\n                    que.push({newX, newY});\n                }\n            }\n        }\n\n        for(int i = 0; i < pos.size(); i++) {\n            int x_ = pos[i][0];\n            int y_ = pos[i][1];\n\n            minDist[index][i] = t[x_][y_];\n        }\n    }\n\n    int solve(vector<vector<int>>& minDist, int idx, int mask, int n, bool Alice, int dp[51][1<<15]) {\n        if(mask == 0) {\n            return 0; \n        }\n\n        if(dp[idx][mask] != -1) {\n            return dp[idx][mask];\n        }\n\n        int result = (Alice == true) ? -1 : 1e9;\n\n        for(int i = 1; i < n; i++) {\n            if(mask & (1 << (i-1))) {\n                int moves = minDist[idx][i];\n\n                if(Alice == true) {\n                    result = max(result, moves + solve(minDist, i, mask ^ (1 << (i-1)), n, !Alice, dp));\n                } else {\n                    result = min(result, moves + solve(minDist, i, mask ^ (1 << (i-1)), n, !Alice, dp));\n                }\n            }\n        }\n\n        return dp[idx][mask] = result;\n    }\n\n    int maxMoves(int kx, int ky, vector<vector<int>>& positions) {\n        int n = positions.size();\n\n        vector<vector<int>> pos;\n        pos.push_back({kx, ky});\n        for(auto &vec : positions) {\n            pos.push_back({vec[0], vec[1]});\n        }\n        vector<vector<int>> minDist(n+1, vector<int>(n+1, 0));\n\n        for(int i = 0; i < pos.size(); i++) {\n            int x = pos[i][0];\n            int y = pos[i][1];\n            BFS(x, y, i, minDist, pos);\n        }\n        bool Alice = true;\n        int dp[51][1 << 15];\n        memset(dp, -1, sizeof(dp));\n        return solve(minDist, 0, (1<<n)-1, pos.size(), Alice, dp);\n    }\n}; \n\nTime Complexity\nBFS Complexity: For each position, we perform a BFS. Since the grid size is fixed (50x50), the BFS has a complexity of O(1). However, since it runs for every position, the effective complexity for all BFS runs is O(n).\nDP Complexity: The bitmask DP has O(n * 2^n) complexity, where n is the number of positions. This is because there are n possible states for each bitmask of size 2^n.\nOverall Time Complexity: O(n * 2^n) due to the DP with bitmasking.
['Dynamic Programming']