POJ 1915 BFS题目

最新推荐文章于 2024-03-02 22:07:34 发布

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最新推荐文章于 2024-03-02 22:07:34 发布

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分类专栏： ACM-搜索文章标签： bfs poj

本文链接：https://blog.youkuaiyun.com/qq_31736627/article/details/52003670

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该博客介绍了如何使用BFS算法解决POJ 1915题，即在棋盘上求解骑士从起点到终点的最短移动步数。通过示例输入和输出，解释了问题背景和解决方案，并强调当起点和终点相同时，步数为0。

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经典的BFS题目，题目的链接如下：
http://poj.org/problem?id=1915

Knight Moves
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 24561 Accepted: 11592
Description

Background
Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him?
The Problem
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov.
For people not familiar with chess, the possible knight moves are shown in Figure 1.

Input

The input begins with the number n of scenarios on a single line by itself.
Next follow n scenarios. Each scenario consists of three lines containing integer numbers. The first line specifies the length l of a side of the chess board (4 <= l <= 300). The entire board has size l * l. The second and third line contain pair of integers {0, …, l-1}*{0, …, l-1} specifying the starting and ending position of the knight on the board. The integers are separated by a single blank. You can assume that the position