矩阵A=A1×A2 ×…… ×An
A[i,j] A[1,n]=A[1,k] ×A[k+1,n] m[i][j] m[i][j]的递
归定义: 0 i=j m[i][j]=min{m[i][k]+ m[k+1][j]+ni-1nknj} i<j
s[i][j]18
12/14/2012 6:50 PM
矩阵A=A1×A2 ×…… ×An
A[i,j] A[1,n]=A[1,k] ×A[k+1,n] m[i][j] m[i][j]的递
归定义: 0 i=j m[i][j]=min{m[i][k]+ m[k+1][j]+ni-1nknj} i<j
s[i][j]18
12/14/2012 6:50 PM