给定一个字符矩阵，其中每个单元格都具有以下值之一。
'C' --> This cell has coin
'#' --> This cell is a blocking cell.
We can not go anywhere from this.
'E' --> This cell is empty. We don't get
a coin, but we can move from here. 初始位置是单元格(0，0)，初始方向是正确的。
以下是跨单元移动的规则。
如果face是Right，那么我们可以移到下面的单元格
- 向前移动一步，即单元格(i，j + 1)和方向保持正确。
- 向下移动一步，并向左移动，即单元格(i + 1，j)和方向变为左。
如果脸是左，那么我们可以移动到下面的单元格
- 向前移动一步，即单元格(i，j-1)和方向保持左。
- 向下移动一步，并向右移动，即单元格(i + 1，j)和方向变为右移。
最终位置可以在任何地方，最终方向也可以是任何东西。目标是收集最多的硬币。
例子：
强烈建议您最小化浏览器，然后自己尝试。
可以按以下方式递归定义以上问题：
maxCoins(i, j, d): Maximum number of coins that can be collected if we begin at cell (i, j) and direction d. d can be either 0 (left) or 1 (right) // If this is a blocking cell, return 0. isValid() checks // if i and j are valid row and column indexes. If (arr[i][j] == '#' or isValid(i, j) == false) return 0 // Initialize result If (arr[i][j] == 'C') result = 1; Else result = 0; If (d == 0) // Left direction return result + max(maxCoins(i+1, j, 1), // Down maxCoins(i, j-1, 0)); // Ahead in left If (d == 1) // Right direction return result + max(maxCoins(i+1, j, 1), // Down maxCoins(i, j+1, 0)); // Ahead in right下面是上述递归算法的C++实现。
C++
// A Naive Recursive C++ program to find maximum number of coins // that can be collected before hitting a dead end #includeusing namespace std; #define R 5 #define C 5 // to check whether current cell is out of the grid or not bool isValid(int i, int j) { return (i >=0 && i < R && j >=0 && j < C); } // dir = 0 for left, dir = 1 for facing right. This function returns // number of maximum coins that can be collected starting from (i, j). int maxCoinsRec(char arr[R][C], int i, int j, int dir) { // If this is a invalid cell or if cell is a blocking cell if (isValid(i,j) == false || arr[i][j] == '#') return 0; // Check if this cell contains the coin 'C' or if its empty 'E'. int result = (arr[i][j] == 'C')? 1: 0; // Get the maximum of two cases when you are facing right in this cell if (dir == 1) // Direction is right return result + max(maxCoinsRec(arr, i+1, j, 0), // Down maxCoinsRec(arr, i, j+1, 1)); // Ahead in right // Direction is left // Get the maximum of two cases when you are facing left in this cell return result + max(maxCoinsRec(arr, i+1, j, 1), // Down maxCoinsRec(arr, i, j-1, 0)); // Ahead in left } // Driver program to test above function int main() { char arr[R][C] = { {'E', 'C', 'C', 'C', 'C'}, {'C', '#', 'C', '#', 'E'}, {'#', 'C', 'C', '#', 'C'}, {'C', 'E', 'E', 'C', 'E'}, {'C', 'E', '#', 'C', 'E'} }; // As per the question initial cell is (0, 0) and direction is // right cout << "Maximum number of collected coins is " << maxCoinsRec(arr, 0, 0, 1); return 0; }
Python3
# A Naive Recursive Python 3 program to # find maximum number of coins # that can be collected before hitting a dead end R= 5 C= 5 # to check whether current cell is out of the grid or not def isValid( i, j): return (i >=0 and i < R and j >=0 and j < C) # dir = 0 for left, dir = 1 for facing right. # This function returns # number of maximum coins that can be collected # starting from (i, j). def maxCoinsRec(arr, i, j, dir): # If this is a invalid cell or if cell is a blocking cell if (isValid(i,j) == False or arr[i][j] == '#'): return 0 # Check if this cell contains the coin 'C' or if its empty 'E'. if (arr[i][j] == 'C'): result=1 else: result=0 # Get the maximum of two cases when you are facing right in this cell if (dir == 1): # Direction is right return (result + max(maxCoinsRec(arr, i+1, j, 0), maxCoinsRec(arr, i, j+1, 1))) # Direction is left # Get the maximum of two cases when you are facing left in this cell return (result + max(maxCoinsRec(arr, i+1, j, 1), maxCoinsRec(arr, i, j-1, 0))) # Driver program to test above function if __name__=='__main__': arr = [ ['E', 'C', 'C', 'C', 'C'], ['C', '#', 'C', '#', 'E'], ['#', 'C', 'C', '#', 'C'], ['C', 'E', 'E', 'C', 'E'], ['C', 'E', '#', 'C', 'E'] ] # As per the question initial cell is (0, 0) and direction is # right print("Maximum number of collected coins is ", maxCoinsRec(arr, 0, 0, 1)) # this code is contributed by ash264
C++
// A Dynamic Programming based C++ program to find maximum // number of coins that can be collected before hitting a // dead end #includeusing namespace std; #define R 5 #define C 5 // to check whether current cell is out of the grid or not bool isValid(int i, int j) { return (i >=0 && i < R && j >=0 && j < C); } // dir = 0 for left, dir = 1 for right. This function returns // number of maximum coins that can be collected starting from // (i, j). int maxCoinsUtil(char arr[R][C], int i, int j, int dir, int dp[R][C][2]) { // If this is a invalid cell or if cell is a blocking cell if (isValid(i,j) == false || arr[i][j] == '#') return 0; // If this subproblem is already solved than return the // already evaluated answer. if (dp[i][j][dir] != -1) return dp[i][j][dir]; // Check if this cell contains the coin 'C' or if its 'E'. dp[i][j][dir] = (arr[i][j] == 'C')? 1: 0; // Get the maximum of two cases when you are facing right // in this cell if (dir == 1) // Direction is right dp[i][j][dir] += max(maxCoinsUtil(arr, i+1, j, 0, dp), // Down maxCoinsUtil(arr, i, j+1, 1, dp)); // Ahead in rught // Get the maximum of two cases when you are facing left // in this cell if (dir == 0) // Direction is left dp[i][j][dir] += max(maxCoinsUtil(arr, i+1, j, 1, dp), // Down maxCoinsUtil(arr, i, j-1, 0, dp)); // Ahead in left // return the answer return dp[i][j][dir]; } // This function mainly creates a lookup table and calls // maxCoinsUtil() int maxCoins(char arr[R][C]) { // Create lookup table and initialize all values as -1 int dp[R][C][2]; memset(dp, -1, sizeof dp); // As per the question initial cell is (0, 0) and direction // is right return maxCoinsUtil(arr, 0, 0, 1, dp); } // Driver program to test above function int main() { char arr[R][C] = { {'E', 'C', 'C', 'C', 'C'}, {'C', '#', 'C', '#', 'E'}, {'#', 'C', 'C', '#', 'C'}, {'C', 'E', 'E', 'C', 'E'}, {'C', 'E', '#', 'C', 'E'} }; cout << "Maximum number of collected coins is " << maxCoins(arr); return 0; } Output:Maximum number of collected coins is 8Time Complexity of above solution is O(R x C x d). Since d is 2, time complexity can be written as O(R x C).Thanks to Gaurav Ahirwar for suggesting above solution.
输出：
Maximum number of collected coins is 8上述解决方案递归的时间复杂度是指数的。我们可以使用动态规划在多项式时间内解决此问题。这个想法是使用3维表dp [R] [C] [k]，其中R是行数，C是列数，d是方向。下面是基于动态编程的C++实现。
C++
// A Dynamic Programming based C++ program to find maximum // number of coins that can be collected before hitting a // dead end #includeusing namespace std; #define R 5 #define C 5 // to check whether current cell is out of the grid or not bool isValid(int i, int j) { return (i >=0 && i < R && j >=0 && j < C); } // dir = 0 for left, dir = 1 for right. This function returns // number of maximum coins that can be collected starting from // (i, j). int maxCoinsUtil(char arr[R][C], int i, int j, int dir, int dp[R][C][2]) { // If this is a invalid cell or if cell is a blocking cell if (isValid(i,j) == false || arr[i][j] == '#') return 0; // If this subproblem is already solved than return the // already evaluated answer. if (dp[i][j][dir] != -1) return dp[i][j][dir]; // Check if this cell contains the coin 'C' or if its 'E'. dp[i][j][dir] = (arr[i][j] == 'C')? 1: 0; // Get the maximum of two cases when you are facing right // in this cell if (dir == 1) // Direction is right dp[i][j][dir] += max(maxCoinsUtil(arr, i+1, j, 0, dp), // Down maxCoinsUtil(arr, i, j+1, 1, dp)); // Ahead in rught // Get the maximum of two cases when you are facing left // in this cell if (dir == 0) // Direction is left dp[i][j][dir] += max(maxCoinsUtil(arr, i+1, j, 1, dp), // Down maxCoinsUtil(arr, i, j-1, 0, dp)); // Ahead in left // return the answer return dp[i][j][dir]; } // This function mainly creates a lookup table and calls // maxCoinsUtil() int maxCoins(char arr[R][C]) { // Create lookup table and initialize all values as -1 int dp[R][C][2]; memset(dp, -1, sizeof dp); // As per the question initial cell is (0, 0) and direction // is right return maxCoinsUtil(arr, 0, 0, 1, dp); } // Driver program to test above function int main() { char arr[R][C] = { {'E', 'C', 'C', 'C', 'C'}, {'C', '#', 'C', '#', 'E'}, {'#', 'C', 'C', '#', 'C'}, {'C', 'E', 'E', 'C', 'E'}, {'C', 'E', '#', 'C', 'E'} }; cout << "Maximum number of collected coins is " << maxCoins(arr); return 0; } 输出：
Maximum number of collected coins is 8上述解决方案的时间复杂度为O(R x C xd)。由于d为2，因此时间复杂度可以写为O(R x C)。
感谢Gaurav Ahirwar提出上述解决方案。