program nonsymmetric
implicit none
integer, parameter :: n = 4 !// depend on your equation
integer :: i, j, mm, tmp, nn, fileid, first, num
real(kind=8) :: matrix(n,n), b(n), x(n)
real(kind=8), allocatable :: aa(:)
integer, allocatable :: ia(:), ja(:)
!// =====================the variables of pardiso====================
integer(kind=8) :: pt(64) !// kind=8:x64; kind=4:win32
integer :: maxfct, mnum, mtype, phase, nrhs, error, msglvl
integer, allocatable :: perm(:)
integer :: iparm(64)
integer, external :: mkl_get_max_threads
!// =====================the variables of pardiso====================
type :: node
real(kind=8) :: s
integer :: k1, k2
type (node), pointer :: next
end type node
type (node), pointer :: head, tail, p, q
!// input data
allocate( perm( n ) )
open ( newunit = fileid, file = 'invM2M1.dat' ) !// The sparse matrix data needs to be given by itself
do i = 1, n
read( fileid,* ) matrix(i,:)
end do
read( fileid,* ) b
close( fileid )
!// =========================store data by CSR format=======================
first = 1
if ( .not.associated(p) ) then
allocate( p )
q => p
nullify( p%next )
q%k2 = first
tmp = q%k2
end if
num = 0 !// calculate the number of no-zero
do i = 1, n
mm = 0 !// calculate the position of no-zero in every row
do j = 1, n
if ( matrix(i,j) /= 0.0 ) then
if ( .not.associated(head) ) then
allocate( head )
tail => head
nullify( tail%next )
tail%s = matrix(i,j)
tail%k1 = j
num = num + 1
mm = mm + 1
else
allocate( tail%next )
tail => tail%next
tail%s = matrix(i,j)
tail%k1 = j
num = num + 1
mm = mm + 1
nullify( tail%next )
end if
end if
end do
if ( associated(p) ) then
allocate( q%next )
q => q%next
q%k2 = tmp + mm
tmp = q%k2
nullify( q%next )
end if
end do
allocate( aa(num), ja(num), ia(n+1) ) !// store the no-zero element
!// output a and ja
tail => head
j = 1
do
if ( .not.associated(tail) ) exit
aa(j) = tail%s
ja(j) = tail%k1
j = j + 1
tail => tail%next
end do
!// output ia
q => p
j = 1
do
if ( .not.associated(q) ) exit
ia(j) = q%k2
j = j + 1
q => q%next
end do
nullify( head, tail, p, q )
!// =========================store data by CSR format=======================
write ( *,'(a)' ) '>> Original data'
do i = 1, n
write( *,'(*(f7.2))' ) matrix(i,:)
end do
write ( *,'(a)' ) '>> CSR data'
write ( *,'(*(f7.2))' ) aa
write ( *,'(a)' ) '>> B data'
write ( *,'(*(f7.2))' ) b
write ( *,'(a)' ) '>> Colnum of CSR data'
write ( *,'(*(i4))' ) ja
write ( *,'(a)' ) '>> The position of CSR data'
write ( *,'(*(i4))' ) ia
!// ===========================solve equations=============================
pt = 0 !// pointer initialization
maxfct = 1; mnum = 1; mtype = 11 !// mtype = -2: symmetric nonpositive definite matrix, mtype = 2: symmetric positive definite matrix
perm = 0; nrhs = 1
x = 0.d0
iparm(1) = 0 !// iparm use default values
iparm(3) = mkl_get_max_threads() !// iparm(3): parallel threads are obtained by MKL. In general, the number of cpus is the same as the number of threads
error = 0
msglvl = 0
phase = 12 !// LU decompose
call pardiso( pt, maxfct, mnum, mtype, phase, n, aa, ia, ja, perm, nrhs, iparm, msglvl, b, x, error )
phase = 33 !// solve equations
call pardiso( pt, maxfct, mnum, mtype, phase, n, aa, ia, ja, perm, nrhs, iparm, msglvl, b, x, error )
write ( *,'(a)' ) '>> The solution as follows ...... '
do i = 1, n
write( *,'(f7.4)' ) x(i)
end do
phase = -1
call pardiso (pt, maxfct, mnum, mtype, phase, n, aa, ia, ja, perm, nrhs, iparm, msglvl, b, x, error)
deallocate( aa, ja, ia, perm )
!// ===========================stop solving equations=============================
end program nonsymmetric
結果如下：
>> Original data
1.00 7.00 0.00 0.00
0.00 2.00 8.00 0.00
5.00 0.00 3.00 9.00
0.00 6.00 0.00 4.00
>> CSR data
1.00 7.00 2.00 8.00 5.00 3.00 9.00 6.00 4.00
>> B data
15.00 28.00 50.00 28.00
>> Colnum of CSR data
1 2 2 3 1 3 4 2 4
>> The position of CSR data
1 3 5 8 10
>> The solution as follows ......
1.0000
2.0000
3.0000
4.0000
請按任意鍵繼續. . .
