标签：acm uva backtracking brute force
题目如下：| Don‘t Get Rooked |
In chess, the rook is a piece that can move any number of squaresvertically or horizontally. In this problem we will consider smallchess boards (at most 4
4)
that can also contain walls through whichrooks cannot move. The goal is to place as many rooks on a board aspossible so that no two can capture each other. A configuration ofrooks is
legal provided that no two rooks are on the samehorizontal row or vertical column unless there is at least one wallseparating them.
The following image shows five pictures of the same board. Thefirst picture is the empty board, the second and third pictures show legalconfigurations, and the fourth and fifth pictures show illegal configurations.For this board, the maximum number of rooks
in a legal configurationis 5; the second picture shows one way to do it, but there are severalother ways.
Your task is to write a program that, given a description of a board,calculates the maximum number of rooks that can be placed on theboard in a legal configuration.
4 .X.. .... XX.. .... 2 XX .X 3 .X. X.X .X. 3 ... .XX .XX 4 .... .... .... .... 0
5 1 5 2 4
跟八皇后问题差不多，就是多了墙而已，这样不一定每行每列只有一个点，中间可能隔着墙。所以用一个函数判断两点之间是否有墙。用一个函数判断一个点是否能被放置，能够的条件是它与所在行所在列的任何一个已经放置的点之间都有墙。有了这个函数后就可以对每个点展开DFS，更新最大值。
AC的代码如下：
UVA Don't Get Rooked,布布扣,bubuko.com
标签：acm uva backtracking brute force
原文地址：http://blog.csdn.net/u013840081/article/details/37829283
