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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;
    }
};

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原文：https://www.cnblogs.com/babydragon/p/11582343.html