"4Blocks" is a two player cooperative game played on a special
board. The board is a grid composed of 1x1 square cells. There are
two different kinds of blocks: '1' blocks and '4' blocks. '1' blocks
are 1x1, and '4' blocks are 2x2:
You must place blocks on the board so that their sides are aligned
to the grid lines and no two blocks overlap. The final score is the
sum of the values in each cell. '1' blocks are worth 1 point, and
'4' blocks are worth 16 points because they cover 4 cells and each
cell is worth 4 points.
Your friend has taken his turn and
placed a number of '1' blocks on the board. The current
configuration is given in the vector <string> grid. The
j-th character of the i-th element of grid is '.' if the cell
at row i, column j is empty, and '1' if your friend placed a '1'
block in that cell. It is now your turn, and you can place any
number of '1' or '4' blocks on the board, but you cannot remove any
of the blocks that have already been placed. Return the maximum
score that can be achieved. For example, the following images show
one possible starting state, and the optimal placement of blocks
from that state:
The
final score would be 4*16 + 6*1 = 70
想法：求解最大的得分，也就是求出在这些空格中，可以摆放的最大的blocks的个数c，然后 16*c + n*m - 4*c既是其最大的得分。采用动态规划求解，设dp[x][y][s]表示在从方格(x,y)开始，y列的前一列的摆放状态为s,按列遍历到最后，能够摆放blocks最多有多少个。
转移方程：(x,y)的摆放状态只与前一列相关。如果（x,y）被blocks覆盖，即判断前一列状态s&(1<<x)是否为真，如果为真，则说明（X,Y）被覆盖，则dp[x][y][s] = dp[x+1][y][mask], mask = s ^(1<<x); 判断(x,y)能否摆放blocks，如果不能，dp[x][y][s] = dp[x+1][y][s],如果能，则 dp[x][y][s] = max{dp[x+1][y][s],dp[x+2][y][s'] s' = s | (1<<x) | (1<<(x+1))}
//Yings 代码
int a[30][30][1<<10];
class FourBlocks{
public:
int n,m;
vector<string> g;
int dp(int x,int y,int mask)
{
int &ret = a[x][y][mask];
if(ret >= 0) return ret;
ret = 0;
if(y == m-1) return ret = 0;
if(x == n) return ret = dp(0,y+1,mask);
int nm = mask;
if(mask & (1 << x))
{
nm ^= (1<<x);
return ret = dp(x+1,y,nm);
}
if(x == n-1 || g[x][y] == '1' || g[x][y+1] == '1')
return ret = dp(x+1,y,nm);
if(g[x+1][y]=='1' || g[x+1][y+1] == '1' || (mask & (1<<(x+1))))
return ret = dp(x+1,y,nm);
ret = dp(x+1,y,nm);
int temp = 1 + dp(x+2,y,nm | (1<<x) | (1<<(x+1)));
if(temp > ret)
ret = temp;
return ret;
}
int maxScore(vector<string> grid)
{
g = grid;
n = (int)g.size();
m = (int)g[0].size();
memset(a,0xff,sizeof(a));
int c = dp(0,0,0);
printf("%d\n",c);
return 12 * c + n*m;
}
};
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