POJ 2411 Mondriaan's Dream(压缩DP)
Mondriaan's Dream
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 8974 | Accepted: 5187 |
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
压缩DP;
这题想了好久。
横的用11表示,竖的用法01表示。
合法的状态应该是相邻两行的或是全1,与的话不存在连续的1的个数是奇数的。
/* * POJ 2411 * 状态压缩DP * 一个h*w的矩阵(1<=h,w<=11),只能放1*2的模块,问完全覆盖的不同放发有多少种? * 横着的定义为11,竖着的定义为01, * 然后按行DP,两行状态相或要全为1.两行相与要没有连续的1的个数是奇数个 */ #include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> using namespace std; bool check(int s)//判断s有没有奇数个连续的1 { int ret=0; while(s) { if(s&1)ret++; else { if(ret&1)return false; ret=0; } s>>=1; } if(ret&1)return false; return true; } long long dp[12][2050]; int main() { int n,m; while(scanf("%d%d",&n,&m)==2) { if(n==0 && m==0)break; int tot=(1<<m); memset(dp,0,sizeof(dp)); for(int i=0;i<tot;i++) if(check(i)) dp[1][i]=1; for(int i=1;i<n;i++) for(int j=0;j<tot;j++) if(dp[i][j]!=0) { for(int k=0;k<tot;k++) if( (j|k)==tot-1 && check(j&k) ) dp[i+1][k]+=dp[i][j]; } printf("%I64d\n",dp[n][tot-1]); } return 0; }
人一我百!人十我万!永不放弃~~~怀着自信的心,去追逐梦想