2888 二维RMQ

HDU2888二维RMQ dp[row][col][i][j] 表示[row,row+2^i-1]x[col,col+2^j-1] 二维区间内的最小值 这是RMQ-ST算法的核心: 倍增思想 == min( [row,row+ 2^(i-1)-1]x[col,col+2^j-1], [row+2^(i-1),row+2^i-1]x[col,col+2^j-1] ) = min(dp[row][col][i-1][j], dp[row+(1<<(i-1))][col][i-1][j] ) //y轴不变,x轴二分 （i!=0) 或 == min( [row,row+2^i-1]x[col,col+2^(j-1)-1], [row,row+2^i-1]x[col+2^(i-1),col+2^j-1] ) = min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1] ) //x轴不变,y轴二分 (j!=0) 即: dp[row][col][i][j] = min(dp[row][col][i-1][j], dp[row + (1<<(i-1))][col][i-1][j] ) 或 = min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1] ) 查询[x1,x2]x[y1,y2] 令 kx = (int)log2(x2-x1+1); ky = (int)log2(y2-y1+1); 查询结果为 m1 = dp[x1][y1][kx][ky] = dp[x1][y1][kx][ky]; m2 = dp[x2-2^kx+1][y1][kx]ky] = dp[x2-(1<<kx)+1][y1][kx][ky]; m3 = dp[x1][y2-2^ky+1][kx][ky] = dp[x1][y2-(1<<ky)+1][kx][ky]; m4 = dp[x2-2^kx+1][y2-2^ky+1][kx][ky] = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky]; 结果 = min(m1,m2,m3,m4)