题目：http://poj.org/problem?id=2411
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.
题意：一个矩阵，只能放1*2的木块，问将这个矩阵完全覆盖的不同放法有多少种。
思路:一道很经典的状态dp,但是还是很难想，横着放定义为11，竖着放定义为01.
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 using namespace std; 6 __int64 d[12][1<<12], f[1<<12];//用———int64,因为有可能数很大 7 int h, w, full; 8 9 bool ok(int x) //判断一行里是否出现连续奇数个1，第一行的话不允许 10 { 11 int sum = 0; 12 while(x>0) 13 { 14 if((x&1)==1) 15 sum++; 16 else 17 { 18 if((sum&1)==1) 19 return false; 20 sum = 0; 21 } 22 x = (x>>1); 23 } 24 if((sum&1)==1) 25 return false; 26 return true; 27 } 28 bool check(int x1, int x2) 29 { 30 int xx = (1<<w)-1; 31 if((x1|x2)==xx&&f[x1&x2])//去掉竖着00和竖着单独11的情况，一定不要忘了单独11的情况 32 return true; 33 return false; 34 } 35 int main() 36 { 37 int i, j, k; 38 memset(f, 0, sizeof(f)); 39 full = (1<<11); 40 41 for(i = 0; i < full; i++) 42 if(ok(i)) 43 f[i] = 1; //记录连续的奇数个1 44 while(~scanf("%d%d", &h, &w)) 45 { 46 if(h==0&&w==0) 47 break; 48 full = (1<<w); 49 memset(d, 0, sizeof(d)); 50 for(i = 0; i < full; i++) 51 if(f[i]) 52 d[0][i] = 1; 53 54 for(k = 1; k < h; k++) 55 for(i = 0; i < full; i++) 56 for(j = 0; j < full; j++) 57 { 58 if(check(i, j)) 59 d[k][i] += d[k-1][j]; 60 } 61 printf("%I64d ", d[h-1][full-1]); //最后一行，而且满1 62 } 63 return 0; 64 }