Mondriaan's Dream
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 16037 | Accepted: 9283 |
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.
![](http://poj.org/images/2411_1.jpg)
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!
![](http://poj.org/images/2411_1.jpg)
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
![](http://poj.org/images/2411_2.jpg)
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
题意:略。思路:略。
待优化。
# include <stdio.h> # include <string.h> # define LL long long const int MAXN = (1<<11)+1; LL dp[MAXN], tmp[MAXN]; bool ok[MAXN]; int a, b, up; bool judge(int i) { while(i) { if(i&1) { i >>= 1; if(!(i&1)) return false; i >>= 1; } else i >>= 1; } return true; } void init() { memset(ok, false, sizeof(ok)); memset(tmp, 0, sizeof(tmp)); for(int i=0; i<up; ++i) if(judge(i)) { ok[i] = true; tmp[i] = 1;//首行必须要相邻的1,或者没1。 } } LL fun() { for(int i=2; i<=a; ++i) { memset(dp, 0, sizeof(dp)); for(int j=0; j<up; ++j) for(int k=0; k<up; ++k) { if(!tmp[k] || (j|k)!=up-1) continue; if(!ok[j&k])//上下两种状态按位与后必须要有两个相邻的1,或者没有1。 continue; dp[j] += tmp[k]; } for(int i=0; i<up; ++i) tmp[i] = dp[i]; } return dp[up-1]; } int main() { while(~scanf("%d%d",&a,&b),a+b) { if((a*b)&1) { puts("0"); continue; } if(a < b) { int c = a; a = b; b = c; } up = 1<<b; init(); printf("%lld\n",fun()); } return 0; }