[POJ] 2411 Mondriaan's Dream

Mondriaan's Dream
Time Limit: 3000MS      Memory Limit: 65536K
Total Submissions: 18903        Accepted: 10779
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.
[我是图]
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
Source
Ulm Local 2000

状压DP，二分图的前身

预处理长度为m的二进制数中是否有连续奇数个0

f[i][j] 第i行，形态为j(2)，的方案数

#include<iostream>
#include<cmath>
using namespace std;
int n,m;
long long f[12][1<<12];
bool jug[1<<12];
int main(){
    while(cin>>n>>m){
        if(!n) return 0;
        for(int i=0;i<1<<m;i++){
            bool ans=0,cnt=0;
            for(int k=0;k<m;k++)
                if((i>>k)&1) ans|=cnt,cnt=0;
                else cnt^=1;
            ans|=cnt;
            jug[i]=!ans;//NO '~'
        }
        f[0][0]=1;
        for(int i=1;i<=n;i++){
            for(int j=0;j<1<<m;j++){
                f[i][j]=0;//
                for(int k=0;k<1<<m;k++){
                    if(jug[k|j]&&!(j&k)){
                        f[i][j]+=f[i-1][k];
                    }
                }
            }
        }
        cout<<f[n][0]<<endl;
    }
    return 0;
}