UVA437-The Tower of Babylon（动态规划基础）

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ACM - 动态规划 - UVA437 The Tower of Babylon

UVA437 The Tower of Babylon 题解初始时给了 \(n\) 种长方体方块,每种有无限个,对于每一个方块,我们可以选择一面作为底.然后用这些方块尽可能高地堆叠成一个塔,要求只有一个方块的底的两条边严格小于另一个方块的底的两条边,这个方块才能堆在另一个上面. 问题的思考在于每种方块有无限个,如果我们直接利用该条件问题会变得比较复杂.其实仔细考虑方块堆叠的要求,会发现这是一个约束很强的条件. 注意到,方块堆叠的要求描述的对象不只是方块本身,更细地说,它应该描述的是方块摆放方式…

[动态规划]UVA437 - The Tower of Babylon

The Tower of Babylon Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story: The babylonian…

Uva437 The Tower of Babylon

https://odzkskevi.qnssl.com/5e1fdf8cae5d11a8f572bae96d6095c0?v=1507521965 Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contes…

UVa437 The Tower of Babylon(巴比伦塔)

题目有n(n<=30)种立方体,每种有无穷多个,摞成尽量高的柱子,要求上面的立方体要严格小于下面的立方体. 原题链接分析顶面的大小会影响后续的决策,但不能直接用d[a][b]来表示,因为可能有多个立方体长宽相等,因此用d[i][j](i<n,j<3)表示第i个立方体的第j条边,为了方便比较大小,边按照从小大到排序. 每增加一个立方体后顶面的长宽都会严格减小,因此是一个DAG图上的最长路问题. AC代码 #include "bits/stdc++.h" using…

【DP】【Uva437】UVA437 The Tower of Babylon

传送门 Description Input Output Sample Input Sample Output Case : maximum height = Case : maximum height = Case : maximum height = Case : maximum height = Hint n<=30. Solution 一眼看出这是个DAG上的DP,一个信息分成三个方块存储.然后开始读入,拓扑,输出,过样例,提交,WA…… 这里的关键是拓扑序应该怎么求.首先我的比较函数写…

UVa 437 The Tower of Babylon（经典动态规划）

传送门 Description Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story: The babylonians had n…