文件名称:juzen
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对于矩阵连乘积的最优计算次序问题，设计算Ai…j ,1≤i≤j≤n，所需的最少数乘次数为m[i,j]，原问题的最优值为m[1,n]。
当i=j时，Ai…j=Ai为单一矩阵，无需计算，因此m[i,i]=0，i=1,2,…,n ；
当i<j时，可利用最优子结构性质来计算m[i,j]。事实上，若计算Ai…j的最优次序在Ak和Ak+1之间断开，i≤k<j，则：m[i,j]=m[i,k]+m[k+1,j]+rirk+1rj+1 。
-Even for the matrix product of the calculation of the optimal order problem, the design of operator Ai ... j, 1 ≤ i ≤ j ≤ n, the required number of at least a few by m [i, j], the original problem of optimal values for m [1 , n]. When i = j when, Ai ... j = Ai for a single matrix, without basis, it is m [i, i] = 0, i = 1,2, ..., n when i <j, to be used to sub-optimal structure the nature of computing m [i, j]. In fact, if the calculation of Ai ... j the optimal order in Ak and Ak+ 1 disconnect between, i ≤ k <j, then: m [i, j] = m [i, k]+ M [k+ 1 , j]+ rirk+ 1rj+ 1.
当i=j时，Ai…j=Ai为单一矩阵，无需计算，因此m[i,i]=0，i=1,2,…,n ；
当i<j时，可利用最优子结构性质来计算m[i,j]。事实上，若计算Ai…j的最优次序在Ak和Ak+1之间断开，i≤k<j，则：m[i,j]=m[i,k]+m[k+1,j]+rirk+1rj+1 。
-Even for the matrix product of the calculation of the optimal order problem, the design of operator Ai ... j, 1 ≤ i ≤ j ≤ n, the required number of at least a few by m [i, j], the original problem of optimal values for m [1 , n]. When i = j when, Ai ... j = Ai for a single matrix, without basis, it is m [i, i] = 0, i = 1,2, ..., n when i <j, to be used to sub-optimal structure the nature of computing m [i, j]. In fact, if the calculation of Ai ... j the optimal order in Ak and Ak+ 1 disconnect between, i ≤ k <j, then: m [i, j] = m [i, k]+ M [k+ 1 , j]+ rirk+ 1rj+ 1.
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juzen.cpp