HDU 1312：Red and Black （广度优先搜索）

最新推荐文章于 2022-01-13 00:13:14 发布

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最新推荐文章于 2022-01-13 00:13:14 发布 · 451 阅读

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#Cpp #广度优先搜索

该博客介绍了如何使用广度优先搜索（BFS）解决HDU 1312编程问题，即在一个红色和黑色方格覆盖的矩形房间里，从黑色方格出发，计算能到达的所有黑色方格的数量。文章讨论了问题的输入输出格式，并给出了解题思路和核心代码片段。

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题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=1312

问题和上一篇博客一样，都是图的搜索问题

上一篇博客采用深度优先搜索，这次采用广度优先搜索

对于这个问题两种搜索算法都能AC，时间空间开销也差不多，实现上也差不多，就是把数据结构改一改，point类的核心算法代码改一改

1312: Red and Black

Time Limit: 2000/1000 MS(Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

There is a rectangular room, covered withsquare tiles. Each tile is colored either red or black. A man is standing on ablack tile. From a tile, he can move to one of four adjacent tiles. But hecan't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach byrepeating the moves described above.

Input

The input consists of multiple data sets. Adata set starts with a line containing two positive integers W and H; W and Hare the numbers of tiles in the x- and y- directions, respectively. W and H arenot more than 20.

There are H more lines in the data set, each of which includes W characters.Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears