poj2411Mondriaan's Dream

Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 11810		Accepted: 6866

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

題意：給你n*m的地板，1*2的地板磚，問將地板鋪滿的方案數

解題思路：剛剛接觸輪廓線dp，這題是比較經典的，1*2的地板磚可以橫放，豎放，還可以不放，我們只需分三種情況討論他的輪廓線狀態，詳細分析參見大白書。

代碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long

LL dp[2][1<<11];
int m;
void update(int t,int a,int b)
{
    if(b&(1<<m))
    dp[t][b^(1<<m)]+=dp[1-t][a];
}
int main()
{
    int n;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)break;
        int t=0;
        memset(dp,0,sizeof(dp));
        dp[0][(1<<m)-1]=1;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                t=1-t;
                memset(dp[t],0,sizeof(dp[t]));
                for(int k=0;k<(1<<m);k++)
                {
                    if(!dp[1-t][k])continue;
                    //bufang
                    update(t,k,k<<1);
                    //shufang
                    if(i&&!((1<<m-1)&k))
                    update(t,k,(k<<1)^1^(1<<m));
                    //hengfang
                    if(j&&!(k&1))
                    update(t,k,(k<<1)^3);
                }
            }
        }
        printf("%lld\n",dp[t][(1<<m)-1]);
    }
    return 0;
}