骨牌摆放方案数n*m（状压DP）

题意：https://www.nitacm.com/problem_show.php?pid=1378

如题。

思路：

从第一行for到最后一行，枚举每一行的所有状态，进行转移，注意答案是dp【最后一行】【0】，因为最后一行是唯一确定的。

https://blog.csdn.net/Tc_To_Top/article/details/43891119

 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
 #include <cstdio>//sprintf islower isupper
 #include <cstdlib>//malloc  exit strcat itoa system("cls")
 #include <iostream>//pair
 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin);
 #include <bitset>
 //#include <map>
 //#include<unordered_map>
 #include <vector>
 #include <stack>
 #include <set>
 #include <string.h>//strstr substr
 #include <string>
 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 #include <cmath>
 #include <deque>
 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 #include <vector>//emplace_back
 //#include <math.h>
 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 //******************
 int abss(int a);
 int lowbit(int n);
 int Del_bit_1(int n);
 int maxx(int a,int b);
 int minn(int a,int b);
 double fabss(double a);
 void swapp(int &a,int &b);
 clock_t __STRAT,__END;
 double __TOTALTIME;
 void _MS(){__STRAT=clock();}
 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
 //***********************
 #define rint register int
 #define fo(a,b,c) for(rint a=b;a<=c;++a)
 #define fr(a,b,c) for(rint a=b;a>=c;--a)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define pr printf
 #define sc scanf
 #define ls rt<<1
 #define rs rt<<1|1
 typedef long long ll;
 const double E=2.718281828;
 const double PI=acos(-1.0);
 //const ll INF=(1LL<<60);
 const int inf=(<<);
 const double ESP=1e-;
 const int mod=(int)1e9+;
 const int N=(int)1e5+;

 ll dp[][N];
 bool ok(int x,int l)
 {
     bool f=;
     int cnt=;
     for(int i=;i<=l-;++i)
     {
         if((x>>i)&)
         {
             if(cnt&)
                 return ;
         }
         else
             cnt++;
     }
     if(cnt&)f=;
     return f;
 }

 int main()
 {
     int n,m;
     while(sc("%d%d",&n,&m),n||m)
     {
         if(n*m%==)
         {
             pr("0\n");
             continue;
         }
         int m_=m;
         m=<<m;
         for(int i=;i<=n;++i)
             for(int j=;j<=m;++j)
                 dp[i][j]=;
         dp[][]=;
         for(int i=;i<=n;++i)
             for(int j=;j<m;++j)
                 for(int k=;k<m;++k)
                     if((j&k)==&&ok(j^k,m_))
                         dp[i][k]+=dp[i-][j];
         pr("%lld\n",dp[n][]);
     }
     return ;
 }

 /**************************************************************************************/

 int maxx(int a,int b)
 {
     return a>b?a:b;
 }

 void swapp(int &a,int &b)
 {
     a^=b^=a^=b;
 }

 int lowbit(int n)
 {
     return n&(-n);
 }

 int Del_bit_1(int n)
 {
     return n&(n-);
 }

 int abss(int a)
 {
     return a>?a:-a;
 }

 double fabss(double a)
 {
     return a>?a:-a;
 }

 int minn(int a,int b)
 {
     return a<b?a:b;
 }