poj 2411 Mondriaan's Dream(状态压缩dP)

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.

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 using namespace std;
 __int64 d[][<<], f[<<];//用———int64,因为有可能数很大
 int h, w, full;

 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
 {
     int sum = ;
     while(x>)
     {
         if((x&)==)
         sum++;
         else
         {
             if((sum&)==)
             return false;
             sum = ;
         }
         x = (x>>);
     }
     if((sum&)==)
             return false;
     return true;
 }
 bool check(int x1, int x2)
 {
     int xx = (<<w)-;
     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
     return true;
     return false;
 }
 int main()
 {
     int i, j, k;
     memset(f, , sizeof(f));
     full = (<<);

     for(i = ; i < full; i++)
     if(ok(i))
     f[i] = ;   //记录连续的奇数个1
     while(~scanf("%d%d", &h, &w))
     {
         if(h==&&w==)
         break;
         full = (<<w);
         memset(d, , sizeof(d));
         for(i = ; i < full; i++)
         if(f[i])
         d[][i] = ;

         for(k = ; k < h; k++)
         for(i = ; i < full; i++)
         for(j = ; j < full; j++)
         {
             if(check(i, j))
             d[k][i] += d[k-][j];
         }
         printf("%I64d\n", d[h-][full-]); //最后一行，而且满1
     }
     return ;
 }