Mondriaan's Dream POJ - 2411

POJ - 2411 

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

题意：求用1*2的小方格填一个n*m的矩形框有多少种填充方法

题解：用二进制表示每一行的状态，看代码

//#include<bits/stdc++.h>
//#include <unordered_map>
//#include<unordered_set>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
#include<climits>
#include<queue>
#include<cmath>
#include<stack>
#include<map>
#include<string>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define MT(a,b) memset(a,b,sizeof(a))
const int INF  =  0x3f3f3f3f;
const int O    =  1e5;
const int mod  =  1e9+7;
const int maxn =  1e2+5;
const double PI  =  3.141592653589;
const double E   =  2.718281828459;


//横放 :1 1
//
//竖放 :0
//     1

int n, m;
LL dp[12][1<<12];
vector<int>Next[1<<12];

void dfs(int list, int first, int next)//查找状态first能转移到下一行的状态next，并记录next
{
    if(list==m)
    {
        Next[first].push_back(next);
        return ;
    }
    if((first&(1<<list))==0) //若first第list列为0，next第list列必为1
        dfs(list+1, first, next|(1<<list));
    else  //若first第list列为1
    {
        dfs(list+1, first, next); //竖放
        if( list<m-1  &&  (first & (1<<(list+1))) )
            dfs(list+2, first, next | (1<<list) | (1<<(list+1))); //横放
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)&&n+m)
    {
        if(n*m%2==1) //奇数
        {
            printf("0\n");
            continue;
        }
        for(int first=0;first<(1<<m);first++)
        {
            Next[first].clear();
            dfs(0, first, 0);
        }
        int Max = (1<<m)-1;
        MT(dp, 0); dp[0][Max] = 1; //第0行相当于一个底，确保第1行既能横放也能竖放
        for(int i=1;i<=n;i++)
        {
            for(int state=0;state<=Max;state++)
            {
                ULL siz = Next[state].size();
                for(int j =0; j<siz; j++)
                    dp[i][ Next[state][j] ] += dp[i-1][state];//状态转移
            }
        }
        printf("%lld\n", dp[n][Max]);
    }
    return 0;
}