Poj 2411 Mondriaan & amp; #39; s Dream (pressure DP)

Poj 2411 Mondriaan & #39; s Dream (pressure DP)

Mondriaan's Dream Time Limit:3000 MS Memory Limit:65536 K Total Submissions:12232 Accepted:7142 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 Z limit? Http://www.bkjia.com/kf/ware/vc/ "target =" _ blank "class =" keylink "> keys" pst "> Sample Input 1 21 31 42 22 32 42 114 110 0 Sample Output 10123514451205 Source Ulm Local 2000

The idea is the same as (in the previous article) SGU 131. The key to this type of question is to learn how to enumerate the status and transfer the status.

#include 
 
  #include 
  
   #include 
   
    #define LL long longusing namespace std;const int N=12;LL dp[N][1<
    
     m)    {        if(!b1 && !b2)  dp[x][s1]+=dp[x-1][s2];        return ;    }    if(!b1 && !b2)  dfs(x,y+1,s1*2+1,s2*2,0,0);    if(!b1)         dfs(x,y+1,s1*2+1,s2*2+1-b2,1,0);    dfs(x,y+1,s1*2+b1,s2*2+1-b2,0,0);}void initial(){    len=1<