迷宫中的老鼠问题是众所周知的回溯问题之一。在这里我们将看到这个问题几乎没有变化。假设给定一个nxn迷宫m。起点为左上角 m[0, 0]，终点为右下角 m[n – 1, n - 1]。一只老鼠被放置在起点。我们的目标是找到一条从起点到终点的路径，可以让老鼠到达目的地。这里老鼠可以跳跃（变体）。现在有一些限制
老鼠可以向右或向下移动。迷宫中单元格中带有 0 表示该单元格是被阻止。非零单元格表示有效路径。单元格内的数字表示大鼠从该单元格可以进行的最大跳跃次数。 ul>算法ratinmazebegin if destination is reached, then print the solution matrix else 1. place the current cell inside the solution matrix as 1 2. move forward or jump (check max jump value) and recursively check if move leads to solution or not. 3. if the move taken from the step 2 is not correct, then move down, and check it leads to the solution or not 4. if none of the solutions in step 2 and 3 are correct, then make the current cell 0. end ifend
示例#include <iostream>#define n 4using namespace std;void dispsolution(int sol[n][n]) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) cout << sol[i][j] << " "; cout << endl; }}bool issafe(int maze[n][n], int x, int y) { //check whether x,y is valid or not // when (x, y) is outside of the maze, then return false if (x >= 0 && x < n && y >= 0 && y < n && maze[x][y] != 0) return true; return false;}bool ratmazesolve(int maze[n][n], int x, int y, int sol[n][n]) { if (x == n - 1 && y == n - 1) { //if destination is found, return true sol[x][y] = 1; return true; } if (issafe(maze, x, y)) { sol[x][y] = 1; //mark 1 into solution matrix for (int i = 1; i <= maze[x][y] && i < n; i++) { if (ratmazesolve(maze, x + i, y, sol)) //move right return true; if (ratmazesolve(maze, x, y + i, sol)) //move down return true; } sol[x][y] = 0; //if the solution is not valid, then make it 0 return false; } return false;}bool solvemaze(int maze[n][n]) { int sol[n][n] = { { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 } }; if (!ratmazesolve(maze, 0, 0, sol)) { cout << "solution doesn't exist"; return false; } dispsolution(sol); return true;}main() { int maze[n][n] = { { 2, 1, 0, 0 }, { 3, 0, 0, 1 }, { 0, 1, 0, 1 }, { 0, 0, 0, 1 } }; solvemaze(maze);}
输出1 0 0 01 0 0 10 0 0 10 0 0 1
以上就是迷宫中的老鼠是否可以进行多步或跳跃？的详细内容。