带有障碍物的1*2铺格子
class Solution {
public:
int domino(int n, int m, vector<vector<int>>& broken) {
int a[10],f[10][256],ans=0,o[256],i,j,k;
memset(a,0,sizeof(a));
for(auto b:broken)a[b[0]]|=1<<b[1];
memset(f,128,sizeof(f));
f[0][(1<<m)-1]=0;
//pre process the number of 1 in [i]
for(i=1;i<1<<m;i++)o[i]=o[i>>1]+(i&1);
for(i=0;i<n;i++)
{
for(j=0;j<1<<m;j++)f[i+1][0]=max(f[i+1][0],f[i][j]);
//vertical put
//因为每种状态都枚举到了,所以竖着插若干个
if(i)
for(j=0;j<1<<m;j++)
for(k=0;k<1<<m;k++)
if(!(j&k)&&!(a[i-1]&k)&&!(a[i]&k))
f[i+1][k]=max(f[i+1][k],f[i][j]+o[k]);
for(j=0;j+1<m;j++)
//j col and j+1 col is empty
//ping fang yi ge
if(!(a[i]>>j&1)&&!(a[i]>>j+1&1))
for(k=0;k<1<<m;k++)
if(!(k>>j&1)&&!(k>>j+1&1))
f[i+1][k|1<<j|1<<j+1]=max(f[i+1][k|1<<j|1<<j+1],f[i+1][k]+1);
}
for(i=0;i<1<<m;i++)ans=max(ans,f[n][i]);
return ans;
}
};