[Dynamic planning] Pku2411--mondriaan ' s Dream

Mondriaan ' s Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3732		Accepted: 2156

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 O F His paper is 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 is in this material, he saw at a glance so he ' ll need a computer to calculate the number of ways to fill t He large rectangle whose dimensions were integer values, as well. Help him, so that he dream won ' t turn into a nightmare!

Input the input contains several test cases. Each test case was made up of the 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 is 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 2
1 3
1 4
2 2
2 3
2 4
2 One
4
each 0 0

Sample Output

1
0
1
2
3
5
144
51205

Source ULM Local 2000 The main idea: give a n*m board, ask how many different schemes are completely covered with 1*2 dominoes. Analysis: Classic state-compression dynamic programming. Both N and M have a maximum of 11, so state compression can be used, i.e. the state of each row is represented by a binary number. The definition F[i,j] indicates the number of scenarios where the first i-1 row has been completely overwritten and the first row has a state of J. Then F[i,j]=∑f[i-1,k]. K can make the i-1 line completely covered by placing a domino, and the line I state is J. This state transfer can be implemented with DFS. Before you begin dynamic planning, you first use DFS to preprocess the first possible state of the row. For the first row, there are only two possible actions: 1, not put. -->dfs (X+1,now SHL 1); 2, Horizontal. -->dfs (X+2,now SHL 2 or 3), when x equals m+1, the description gets a viable State, Inc. (F[1,now]), and when X is greater than M, exit Dfs. The DFS process for state transitions is similar to this, with just one more vertical operation, using bit arithmetic, left to the reader to think, or refer to the program. Codes:const maxn= (1 shl 11)-1; var N,m:int64; I:longint; F:ARRAY[1..11,0..MAXN] of Int64; Procedure init (x,now:longint); Begin IF X=m+1 then Inc (F[1,now]); If X>m then exit; Init (X+1,now SHL 1); Init (X+2,now SHL 2 or 3); End Procedure Dfs (X,last,new:longint); Begin IF X=m+1 then Inc (F[i,new],f[i-1,last]); If X>m then exit; DFS (x+1,last SHL 1 or 1,new SHL 1); DFS (x+1,last shl 1,new SHL 1 or 1); DFS (x+2,last SHL 2 or 3,new SHL 2 or 3); End Procedure main; var t:longint; Begin if M>n then BEGIN t:=n; N:=m; m:=t; End Fillchar (F,sizeof (f), 0); Init (1, 0); For i:=2 to N do Dfs (1,0,0); Writeln (F[n, (1 SHL m)-1]); End Begin READLN (N,M); While N+m<>0 does begin main; READLN (N,M); End End.