poj2411 Mondriaan's Dream【状压DP】

Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 20822		Accepted: 11732

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 oriented, i.e. count symmetrical tilings multiple times.

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

Ulm Local 2000

题意：

给定一个h*w的矩形。将矩形划分成1*2的小格子，有多少种方案。

思路：

考虑用行数作为状态，但是转移下一行时需要上一行的划分状态。

所以我们多开一维用于记录状态。用一个整数表示。第k位是1表示第i行第k列的格子是一个竖着的1*2长方形的上半部分。

那么对于第i+1行的状态j， j&k=0表示没有两个相邻行的相同列的格子都是长方形的上半部分。

j|k的二进制表示中，每一段连续的0都是偶数个。j|k是0的位，要么是j和k该位都是0说明这是一个横着的矩形。

只有这两种情况都满足时，(i, k)才能转移到(i+1, j)。

最后一行要输出的应该是没有一个1的情况。

注意要使用long long

 //#include <bits/stdc++.h>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<stdio.h>
 #include<cstring>
 #include<vector>
 #include<map>
 #include<set>

 #define inf 0x3f3f3f3f
 using namespace std;
 typedef long long LL;

 int h, w;
 const int maxn =  << ;
 bool in_s[maxn];
 long long dp[][maxn];

 int main(){
     while(scanf("%d%d", &h, &w) != EOF && (h || w)){
         for(int i = ; i <  << w; i++){
             bool cnt = , has_odd = ;
             for(int j = ; j < w; j++){
                 if(i >> j & ) has_odd |= cnt, cnt = ;
                 else cnt ^= ;
             }
             in_s[i] = has_odd | cnt ?  : ;
         }

         //memset(dp, 0, sizeof(dp));
         dp[][] = ;
         for(int i = ; i <= h; i++){
             for(int j = ; j <  << w; j++){
                 dp[i][j] = ;
                 for(int k = ; k <  << w; k++){
                     if((k & j) ==  && in_s[k | j]){
                         dp[i][j] += dp[i - ][k];
                     }
                 }
             }
         }
         printf("%lld\n", dp[h][]);
     }
     return ;
 }