题目:Mondriaan's Dream 链接:http://poj.org/problem?id=2411 题意:用 1*2 的瓷砖去填 n*m 的地板,问有多少种填法. 思路: 很久很久以前便做过的一道题目,状压DP,当时写得估计挺艰辛的,今天搜插头DP又搜到它,就先用状压DP写了下,顺利多了,没一会就出来了,可惜因为long long没有1A. 思路挺简单,一行一行解决,每一列用1 表示对下一行有影响,用0 表示对下一行没有影响,所以一行最多2048 种可能,然后要筛选一下,因为有些本身就…

Mondriaan's Dream Time Limit: 3000MS Memory Limit: 65536K 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 d…

[题目分析] 用1*2的牌铺满n*m的格子. 刚开始用到动规想写一个n*m*2^m,写了半天才知道会有重复的情况. So Sad. 然后想到数据范围这么小,爆搜好了.于是把每一种状态对应的转移都搜了出来. 加了点优(gou)化(pi),然后poj上1244ms垫底. 大概的方法就是考虑每一层横着放的情况,剩下的必须竖起来的情况到下一层取反即可. 然后看了 <插头DP-从入门到跳楼> 这篇博客,怒抄插头DP 然后16ms了,自己慢慢YY了一下,写出了风(gou)流(pi)倜(bu)傥(tong)…

题目大意:一个矩阵,只能放1*2的木块,问将这个矩阵完全覆盖的不同放法有多少种. 解析:如果是横着的就定义11,如果竖着的定义为竖着的01,这样按行dp只需要考虑两件事儿,当前行&上一行,是不是全为1,不是说明竖着有空(不可能出现竖着的00),另一个要检查当前行里有没有横放的,但为奇数的1. 原代码链接:http://blog.csdn.net/accry/article/details/6607703 首先我个人感觉,横着是11,竖着是01 这个方法很牛逼,然后就是先预处理ok数组,之后就要判…

状压DP 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…

一.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…

Mondriaan's Dream 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…

题目:http://poj.org/problem?id=2411 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 eac…

题意:有一个n*m的棋盘,要求用1*2的骨牌来覆盖满它,有多少种方案?(n<12,m<12) 思路: 由于n和m都比较小,可以用轮廓线,就是维护最后边所需要的几个状态,然后进行DP.这里需要维护的状态数就是min(n,m).即大概是一行的大小.每次放的时候,只考虑(1)以当前格子为右方,进行横放:(2)以当前格子为下方进行竖放:(3)还有就是可以不放. 3种都是不一样的,所以前面的一种状态可能可以转为后面的几种状态,只要满足了条件.条件是,横放时,当前格子不能是最左边的:竖放时,当前格子不能是…

题目链接: http://poj.org/problem?id=2411 Mondriaan's Dream Time Limit: 3000MSMemory Limit: 65536K 问题描述 Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had…