All Roads Lead to Rome!

All Roads Lead to Rome!

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你站在非负数对网格中的点(x, y) 上,希望到达原点(0, 0) 。在任何一点,你都可以向左或向下移动一步。此外,您永远不允许进入tabu 列表中的任何点(原点永远不在tabu 中)。在这些限制条件下,该函数应将从(x,y) 点到原点(0,0) 的不同路径数相加。


提示 1

这个经典组合问题的受限变体有一个相当简单的递归解法。作为基本情况,从原点(0, 0) 到自身的路径数等于空路径的 1(注意存在的空路径与不存在的路径之间的关键区别!)。如果点(x, y)tabu 列表中,则从该点到原点的路径数等于零。否则,从该点到原点的路径数等于两个相邻点 (x-1, y) 和 (x, y-1) 的路径之和。然而,这种简单的递归分支会产生指数级的可能性,因此执行起来可能会过于缓慢。因此,您应该使用lru_cache 对递归进行记忆,或者更好的方法是...

提示 2

完全不要使用递归,而是建立一个二维列表,其中的条目是各个子问题的解决方案。用两个 for 循环按照一定的顺序填入正确的值,以保证当这些循环到达...

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