Mondriaan's Dream POJ - 2411

Mondriaan's Dream POJ - 2411

可以用状压dp，但是要打一下表。暴力枚举行、这一行的状态、上一行的状态，判断如果上一行的状态能转移到这一行的状态就转移。

状态定义：ans[i][S]表示i行前已经全部填满，i行已经填上的列为集合S。如果有竖着的，全部当做用这一行的去补满上一行缺的。

（貌似还是插头dp的入门题）

 #include<cstdio>
 #include<cstring>
 typedef long long LL;
 LL f[][];
 LL h,w;
 /*LL ans[12][2050];
 LL h,w;
 bool judge(LL a,LL b)
 {
     LL i,num=0,t1,t2;
     for(i=1;i<=w;i++)
     {
         t1=a&1;
         t2=b&1;
         a>>=1;
         b>>=1;
         if(t1==1&&t2==1)
             num++;
         else if(t1==0&&t2==0)
             return false;
         else
         {
             if(num&1)
                 return false;
             num=0;
         }
     }
     if(num&1)
         return false;
     else
         return true;
 }
 LL get(LL h,LL w)
 {
     LL i,j,k;
     memset(ans,0,sizeof(ans));
     ans[0][(1<<w)-1]=1;
     for(i=1;i<=h;i++)
         for(j=0;j<(1<<w);j++)
             for(k=0;k<(1<<w);k++)
                 if(judge(k,j))
                 {
                     ans[i][j]+=ans[i-1][k];
                 }
     return ans[h][(1<<w)-1];
 }
 int main()
 {
     for(h=1;h<=11;h++)
         for(w=1;w<=11;w++)
             printf("f[%lld][%lld]=%lld;\n",h,w,get(h,w));
     return 0;
 }*/
 int main()
 {
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     scanf("%lld%lld",&h,&w);
     while(h!=&&w!=)
     {
         printf("%lld\n",f[h][w]);
         scanf("%lld%lld",&h,&w);
     }
 }