poj2411 Mondriaan's Dream[简单状压dp]

$11*11$格子板上铺$1*2$地砖方案。以前做过？权当复习算了，毕竟以前学都是浅尝辄止的。。常规题，注意两个条件：上一行铺竖着的则这一行同一位一定要铺上竖的，这一行单独铺横的要求枚举集合中出现连续偶数个的1，预处理一下即可。注意数据及时reset。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<queue>
 7 #define dbg(x) cerr<<#x<<" = "<<x<<endl
 9 using namespace std;
10 typedef long long ll;
11 template<typename T>inline char MIN(T&A,T B){return A>B?A=B,1:0;}
12 template<typename T>inline char MAX(T&A,T B){return A<B?A=B,1:0;}
13 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
14 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
15 template<typename T>inline T read(T&x){
16     x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1;
17     while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
18 }
19 const int N=12;
20 ll f[N][1<<N-1];
21 int n,m,ful,g[1<<N-1];
22 
23 int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
24     for(register int i=0;i<(1<<11);++i){
25         int cnt=0;
26         for(register int j=0;j<11;++j){
27             if(!((1<<j)&i)){if(cnt&1)break;cnt=0;}
28             else ++cnt;
29         }
30         if(!(cnt&1))g[i]=1;
31     }
32     while(read(n),read(m),n){
33         if((n*m)&1){printf("0\n");continue;}
34         f[0][ful]=0;ful=(1<<m)-1;f[0][ful]=1;
35         for(register int i=1;i<=n;++i){
36             for(register int j=0;j<=ful;++j){
37                 f[i][j]=0;
38                 for(register int k=0;k<=ful;++k)if(((j|k)==ful)&&(g[j&k])){
39                     f[i][j]+=f[i-1][k];
40                 }
41             }
42         }
43         printf("%lld\n",f[n][ful]);
44     }
45     return 0;
46 }