使用BFS算法完成以下问题:Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards. You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.) Input First line contains two integers stand for N and M. Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend. Process to the end of the file. Output For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."
时间: 2024-04-28 19:22:58 浏览: 9
解题思路:
这道题可以使用 BFS 算法来解决。首先需要找到起点和终点,并标记出所有的墙和警卫。使用 BFS 算法搜索整个迷宫,每次向四个方向扩展,如果扩展到的位置是墙或者警卫则跳过,否则加入到队列中继续扩展。如果扩展到终点则直接返回步数。最后如果没有找到终点,则说明无法到达,输出 "Poor ANGEL has to stay in the prison all his life."。
具体实现可以按照以下步骤进行:
1. 读入输入数据,找到起点和终点,并标记出所有的墙和警卫。
2. 使用 BFS 算法搜索整个迷宫,每次向四个方向扩展,如果扩展到的位置是墙或者警卫则跳过,否则加入到队列中继续扩展。
3. 如果扩展到终点则直接返回步数。
4. 最后如果没有找到终点,则输出 "Poor ANGEL has to stay in the prison all his life."。
代码实现:
相关问题
Cycle pruning: Check whether a node is in the path already before adding the path to Frontier.请解释一下
这句话的意思是,在进行基于图的搜索算法时,比如广度优先搜索(BFS)或者深度优先搜索(DFS)时,有时候会出现环路的情况,即搜索到达一个已经访问过的节点,此时如果不进行处理,会导致算法陷入死循环。
为了避免这种情况,可以在将路径添加到搜索的边界(Frontier)之前,先检查路径上是否已经包含了当前节点。如果已经包含,则该路径会形成一个环路,需要将其丢弃,否则可以将其加入到搜索的边界中。这样可以保证搜索算法不会陷入死循环,同时也能够避免重复访问已经访问过的节点,提高搜索效率。
Cycle pruning(循环修剪)是一种常用的优化方法,主要应用于图搜索和神经网络权重剪枝等领域。在图搜索中,循环修剪可以避免算法陷入死循环,提高搜索效率;在神经网络中,循环修剪可以去除冗余权重,减小模型大小和计算复杂度,提高模型效率和通用性。
对比一下Dijkstra算法与BFS算法以及A*算法
Dijkstra算法、BFS算法和A*算法都是图搜索算法。Dijkstra算法是一种单源最短路径算法,它能够找到从起点到所有其他节点的最短路径。BFS算法是一种广度优先搜索算法,它能够找到从起点到目标节点的最短路径。A*算法是一种启发式搜索算法,它结合了Dijkstra算法和BFS算法的优点,能够更快地找到最短路径。在实际应用中,选择哪种算法取决于具体的问题和数据结构。