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以下分析TComDataCU.cpp。这个cpp是很重要的,要分几次分析完,这是分析TComDataCU.cpp(一)。
Void TComDataCU::getPartPosition( UInt partIdx, Int& xP, Int& yP, Int& nPSW, Int& nPSH){ UInt col = m_uiCUPelX; UInt row = m_uiCUPelY; switch ( m_pePartSize[0] ) { case SIZE_2NxN: nPSW = getWidth(0); nPSH = getHeight(0) >> 1; xP = col; yP = (partIdx ==0)? row: row + nPSH; break; case SIZE_Nx2N: nPSW = getWidth(0) >> 1; nPSH = getHeight(0); xP = (partIdx ==0)? col: col + nPSW; yP = row; break; case SIZE_NxN: nPSW = getWidth(0) >> 1; nPSH = getHeight(0) >> 1; xP = col + (partIdx&0x1)*nPSW; yP = row + (partIdx>>1)*nPSH; break; case SIZE_2NxnU: nPSW = getWidth(0); nPSH = ( partIdx == 0 ) ? getHeight(0) >> 2 : ( getHeight(0) >> 2 ) + ( getHeight(0) >> 1 ); xP = col; yP = (partIdx ==0)? row: row + getHeight(0) - nPSH; break; case SIZE_2NxnD: nPSW = getWidth(0); nPSH = ( partIdx == 0 ) ? ( getHeight(0) >> 2 ) + ( getHeight(0) >> 1 ) : getHeight(0) >> 2; xP = col; yP = (partIdx ==0)? row: row + getHeight(0) - nPSH; break; case SIZE_nLx2N: nPSW = ( partIdx == 0 ) ? getWidth(0) >> 2 : ( getWidth(0) >> 2 ) + ( getWidth(0) >> 1 ); nPSH = getHeight(0); xP = (partIdx ==0)? col: col + getWidth(0) - nPSW; yP = row; break; case SIZE_nRx2N: nPSW = ( partIdx == 0 ) ? ( getWidth(0) >> 2 ) + ( getWidth(0) >> 1 ) : getWidth(0) >> 2; nPSH = getHeight(0); xP = (partIdx ==0)? col: col + getWidth(0) - nPSW; yP = row; break; default: assert ( m_pePartSize[0] == SIZE_2Nx2N ); nPSW = getWidth(0); nPSH = getHeight(0); xP = col ; yP = row ; break; }}
在 2N*N, 2N*nU,2N*nD 中,上面部分的partIdx 为0,下面部分为1;
在N*2N,nL*2N,nR*2N中, 左边部分的partIdx为0,右边部分为1;
在N*N中, 按四叉树的结构,partIdx 分别为0,1,2,3;
在2N*2N中, partIdx只能为0;
xP,yP的值是CU块左上角的绝对位置值,nPSW,nPSH是当前CU块的宽度和高度,记住,CU是可以嵌套的,CU包含CU,所以有xP,yP,nPSW,nPSH就能够唯一的确定CU块,
限于时间关系,今天先分析至此。