HDU5731 Solid Dominoes Tilings 状压dp+状压容斥

题意：给定n,m的矩阵，就是求稳定的骨牌完美覆盖,也就是相邻的两行或者两列都至少有一个骨牌

分析：第一步：

如果是单单求骨牌完美覆盖，请先去学基础的插头dp(其实也是基础的状压dp)骨牌覆盖

hihocoder有全套课程：骨牌覆盖（一, 二，三），状态压缩（二）

学好了以后，首先打一个预处理没有限制的表，由于赛后补题，我就没自己打，直接从网上粘的表

我的表来自：http://blog.csdn.net/u012015746/article/details/51971977

第二步：

这就是容斥的过程了，我们可以枚举每种列分割状态，计算出每种列分割状态下行合法的方案数

然后用总数，减去一个列分割数为1的，加上列分割数为2.........这就是容斥的过程了

细节处理：每种列状态下如何求合法的行状态呢，每种状态在用一下递推一下就好了（枚举依据是前j行无行分割，后i-j行有行分割）

其实是枚举的第一个行分割线出现的位置，上面不能有，下面就可以随意了

详情请参考上面的链接

复杂度：O（T*(n^2)*(2^m))，大概是这个复杂度

#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 1e9+;
LL RES[][];
void init();
LL dp[];
LL solve(int n,int m){
  int pos[],tot=(<<(m-));
  LL ret=;
  for(int cur=;cur<tot;++cur){
     int cnt=;pos[cnt]=;
     for(int i=;i<m-;++i)if(cur&(<<i))pos[++cnt]=i+;
     pos[++cnt]=m;
     for(int i=cnt;i>;--i)pos[i]-=pos[i-];
     for(int i=;i<=n;++i){
        for(int j=;j<i;++j){
           LL tmp=;
           for(int k=;k<=cnt;++k)
             tmp=tmp*RES[i-j][pos[k]]%mod;
           if(!j)dp[i]=tmp;
           else dp[i]=(dp[i]-tmp*dp[j]%mod+mod)%mod;
        }
     }
     if(cnt&)ret=(ret-dp[n]+mod)%mod;
     else ret=(ret+dp[n])%mod;
  }
  return ret;
}
int main(){
  init();
  int n,m;
  while(~scanf("%d%d",&n,&m)){
     printf("%I64d\n",solve(n,m));
  }
  return ;
}
void init()
{
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}