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Re:DP解题思路：

Posted by tasty at 2016-05-06 08:46:25 on Problem 1071
In Reply To:DP解题思路： Posted by:alpc56 at 2009-08-23 16:46:00

> 按照chase[N-1]到chase[0] DP
> for(k=n-1;k>=0;k--)
> dp[i][j](bool) 表示点坐标(i,j)作为完成后k个chase的起点的可行性
> DP完chase[0] 检查dp[][]得到可行解的个数就OK了
> 真不明白搜索是怎么写得？
三维状态没有爆掉 10000K 内存？  最好还是二维吧。。。

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Re:DP解题思路： (44) tasty 2016-05-06 08:47:39

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