POJ2411Mondriaan's Dream题解动态规划DP

POJ 2411 Mondriaan's Dream (轮廓线DP代码详解) Mondriaan’s Dream(POJ 2411) Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 18772 Accepted: 10717 Description Squares and rectangles fascinated the famous Dutch painter Piet... 阅读详情
Mondriaan's Dream
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 4978 Accepted: 2873

Description

Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways.

Expert as he was in this material, he saw at a glance that he'll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won't turn into a nightmare!

Input

The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.

Output

For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.

Sample Input

1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0

Sample Output

1
0
1
2
3
5
144
51205

Source

状态定义等见:
感谢POJ的Sayakiss学习了他的代码,让我对状态压缩DP有了初步的了解。
下面的代码就是学习他的代码的产物,其实基本一样
POJ 2411 Mondriaan's Dream详细解题报告 里面有非常详细的对于POJ 2411的解题报告,相信对于初学动态规划和深度优先搜索的同学来说有很好的帮助作用。 立即下载

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Mondriaan's Dream POJ - 2411 (状压DP)

题目链接: Mondriaan's Dream POJ - 2411 题目大意:就是给你一个n*m的矩阵,然后问你用2*1的矩阵和1*2的矩阵,有多少种方法能凑出来n*m的矩阵,在不会有重叠的前提下。 具体思路:状压dp,对于1*2的矩阵,我们将这个方块构成的两块都标记为1;对于2*1的矩阵,我们将上面那个方块标记为0,下面那个方块标记为1.然后在具体判断的过程中,先判断第一...

diaomeijiao3430的博客 141

【题解】poj2411 Mondriaan's Dream 状压DP

题目链接 Description Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his ‘toilet series’ (where he had to use his toilet paper to draw...

翻过这座山,他们就会听到你的故事 371

[poj 2411] Mondriaan's Dream{状态压缩DP}

题目 http://poj.org/problem?id=2411 解题思路 代码 #include&lt;cstdio&gt; #include&lt;cstring&gt; #include&lt;algorithm&gt; using namespace std; int n,m; long long f[12][1&lt;&lt;11]; bool s[1&lt;&lt;11]; ...

Nowed 435

Mondriaan‘s Dream 【POJ2411】 题解

行号(枚举顺序)压缩后的状态方案数于是我们可以定义fijxfij​x表示第iii行的状态为jjj(假设前i−1i - 1i−1行都已确定)时的方案数。由于一共有mmm行,我们可以将数组略定大一点,最终求到fm10fm10​的值,便是包括到mmm行的方案总数。截止到上一行的方案数即为上一行所有合法的情况数总和。

IOLOand32sn的博客 335

【poj2411】Mondriaan's Dream 状态压缩dp

AC传送门:http://vjudge.net/problem/POJ-2411 【题目大意】 有一个W行H列的广场,需要用1*2小砖铺盖,小砖之间互相不能重叠,问有多少种不同的铺法? 【题解】 对于每一行有w个位置,所以每一行都有0~2w-1种状态。 对于当前行的状态s,它是由前一行的状态s’转化过来的,显然,对于该行某个位置j: 如果前一行该位置为0,那么该位置可...

anzhi7214的博客 146
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