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给大家提供一个简单一些的方法。。

Posted by 11119999 at 2006-08-17 00:18:07 on Problem 1230
可以考虑直接保存整个stage为一个二维数组,在二维数组中每个墙从左端到右端编号为1-N(N为墙的长度),这样从右向左扫描各列的时候,只要选取该列各行中序号最大的stage[i][j],然后修改该行从j列开始到j-stage[i][j]+1的数值为0就可以了,还有就是注意墙的两个端点先输入哪个都行,其他的没什么了。呵呵。

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