POJ 2411Mondriaan's Dream
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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!
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 21 31 42 22 32 42 114 110 0
Sample Output
10123514451205
采用2进制保存每个点的情况,下一层总是由上一层的状态递推而来。
然后通过搜索来判断横向的方块如何放置。
#include<stdio.h>#include<algorithm>#include<iostream>using namespace std;const int maxn=12;int n,m,N;long long f[maxn][1<<maxn];int ins[1<<maxn];int c[maxn]={3,6,12,24,48,96,192,384,768,1536,3072,6144};long long dfs(int x,int y){long long tot=0;for (int k=0;k<maxn&&c[k]<=y;k++)if ((c[k]&y)==c[k]&&ins[y-c[k]]==0){ins[y-c[k]]=1;tot+=dfs(x,y-c[k]);}return tot+f[x][y];}int main(){while (~scanf("%d%d",&n,&m),n+m){memset(f,0,sizeof(f));m=(1<<m)-1;f[1][0]=1;for (int i=1;i<=n;i++)for (int j=m;j>=0;j--) memset(ins,0,sizeof(ins)),f[i+1][j^m]+=f[i][j]=dfs(i,j);cout<<f[n][m]<<endl;}}
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