poj 2411 Mondriaan's Dream

铺砖问题+dfs+状态压缩

Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 11507		Accepted: 6694

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

Sample Output

1
0
1
2
3
5
144
51205
题意：用1*2的砖去铺h，w的格子，可以横放，也可以竖放。

如果格子h*w为奇数则一定铺不满。

1代表铺，0代表不铺，不铺则下一层一定要铺，抓住下一层一定要把上一层铺满。

第i行只和i-1行有关
枚举i-1行的每个状态，推出由此状态能达到的i行状态
如果i-1行的出发状态某处未放，必然要在i行放一个竖的方块，所以我对上一行状态按位取反之后的状态就是放置了竖方块的状态。
然后用搜索扫一道在i行放横着的方块的所有可能，并且把这些状态累加上i-1的出发状态的方法数，如果该方法数为0，直接continue。
举个例子
2 4
1111
1111
状态可以由
1100 0000 0110 0011 1111
0000 0000 0000 0000 0000
这五种i-1的状态达到，故2 4 的答案为5

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int h,w;
__int64 dp[20][2148],add;
void  dfs(int i,int s,int x)
{
    if(x==w)
        {
            //printf("dpi=%d
",dp[i][s]);
          dp[i][s]+=add;
        //  printf("i=%d,s=%d,dp=%d
",i,s,dp[i][s]);
          return;
        }
        dfs(i,s,x+1);
        if(x<=w-2&&!(s&1<<x)&&!(s&1<<(x+1)))//连续2为不为1.说明上一层没有竖放的。
           dfs(i,s|1<<x|1<<(x+1),x+2);//要并起来。
}
int main()
{
      int i,j,k,s;
    while(~scanf("%d%d",&h,&w))
    {
        memset(dp,0,sizeof(dp));
        if(h==0&&w==0)
             break;
          if((h*w)%2==1)
             {
               printf("0
");
                continue;
             }
            add=1;
          dfs(1,0,0);
          for(i=2;i<=h;i++)
          {
             for(j=0;j<1<<w;j++)
             {
                 if(dp[i-1][j])
                      {
                          add=dp[i-1][j];
                          //printf("add=%d
",add);
                      }
                 else
                    continue;
                dfs(i,~j&((1<<w)-1),0);//&保证有w位数，比如11可以保证有00011.

             }
          }
        printf("%I64d
",dp[h][(1<<w)-1]);
    }
    return 0;
}