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178 lines
5.0 KiB
Fortran
178 lines
5.0 KiB
Fortran
C
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C Parallel Sparse BLAS v2.0
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C (C) Copyright 2006 Salvatore Filippone University of Rome Tor Vergata
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C Alfredo Buttari University of Rome Tor Vergata
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C
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C Redistribution and use in source and binary forms, with or without
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C modification, are permitted provided that the following conditions
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C are met:
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C 1. Redistributions of source code must retain the above copyright
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C notice, this list of conditions and the following disclaimer.
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C 2. Redistributions in binary form must reproduce the above copyright
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C notice, this list of conditions, and the following disclaimer in the
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C documentation and/or other materials provided with the distribution.
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C 3. The name of the PSBLAS group or the names of its contributors may
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C not be used to endorse or promote products derived from this
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C software without specific written permission.
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C
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C THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
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C ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED
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C TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
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C PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE PSBLAS GROUP OR ITS CONTRIBUTORS
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C BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
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C CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
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C SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
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C INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
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C CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
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C ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
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C POSSIBILITY OF SUCH DAMAGE.
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C
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C
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SUBROUTINE ISR(N,X)
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C
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C Quicksort.
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C Adapted from a number of sources, including Don Knuth's TAOCP.
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C
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C .. Scalar Arguments ..
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INTEGER N
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C ..
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C .. Array Arguments ..
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INTEGER X(N)
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C ..
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C .. Local Scalars ..
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INTEGER I, J, XX, ILX, IUX, ISTP, PIV, LPIV
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INTEGER IT1, N1, N2
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INTEGER MAXSTACK,NPARMS,ITHRS
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PARAMETER (MAXSTACK=64,NPARMS=3,ITHRS=16)
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INTEGER ISTACK(NPARMS,MAXSTACK)
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C ..
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C
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C Small inputs will only get through insertion sort.
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C
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IF (N.GT.ITHRS) THEN
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C
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C Init stack pointer
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C
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ISTP = 1
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ISTACK(1,ISTP) = 1
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ISTACK(2,ISTP) = N
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DO WHILE (ISTP.GT.0)
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ILX = ISTACK(1,ISTP)
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IUX = ISTACK(2,ISTP)
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ISTP = ISTP - 1
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c$$$ write(0,*) 'Debug 1: ',ilx,iux
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C
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C Choose a pivot with median-of-three heuristics, leave it
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C in the LPIV location
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C
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I = ILX
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J = IUX
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LPIV = (I+J)/2
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PIV = X(LPIV)
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IF (PIV.LT.X(I)) THEN
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IT1 = X(I)
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X(I) = X(LPIV)
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X(LPIV) = IT1
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PIV = X(LPIV)
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ENDIF
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IF (PIV.GT.X(J)) THEN
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IT1 = X(J)
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X(J) = X(LPIV)
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X(LPIV) = IT1
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PIV = X(LPIV)
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ENDIF
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IF (PIV.LT.X(I)) THEN
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IT1 = X(I)
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X(I) = X(LPIV)
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X(LPIV) = IT1
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PIV = X(LPIV)
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ENDIF
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C
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C Now PIV is correct; place it into first location
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IT1 = X(I)
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X(I) = X(LPIV)
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X(LPIV) = IT1
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I = ILX - 1
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J = IUX + 1
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130 CONTINUE
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I = I + 1
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XK = X(I)
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IF (XK.LT.PIV) GOTO 130
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C
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C Ensure finite termination for next loop
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C
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IT1 = XK
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X(I) = PIV
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140 CONTINUE
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J = J - 1
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XK = X(J)
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IF (XK.GT.PIV) GOTO 140
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X(I) = IT1
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150 CONTINUE
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IF (J.GT.I) THEN
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IT1 = X(I)
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X(I) = X(J)
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X(J) = IT1
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GO TO 130
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END IF
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if (i.eq.ilx) then
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if (x(i).ne.piv) then
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write(0,*) 'Should never ever get here????!!!!'
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stop
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endif
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i = i + 1
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endif
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N1 = (I-1)-ILX+1
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N2 = IUX-(I)+1
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IF (N1.GT.N2) THEN
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if (n1.gt.ithrs) then
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ISTP = ISTP + 1
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ISTACK(1,ISTP) = ILX
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ISTACK(2,ISTP) = I-1
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endif
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if (n2.gt.ithrs) then
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ISTP = ISTP + 1
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ISTACK(1,ISTP) = I
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ISTACK(2,ISTP) = IUX
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endif
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ELSE
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if (n2.gt.ithrs) then
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ISTP = ISTP + 1
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ISTACK(1,ISTP) = I
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ISTACK(2,ISTP) = IUX
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endif
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if (n1.gt.ithrs) then
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ISTP = ISTP + 1
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ISTACK(1,ISTP) = ILX
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ISTACK(2,ISTP) = I-1
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endif
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ENDIF
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ENDDO
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ENDIF
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DO J=N-1,1,-1
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IF (X(J+1).LT.X(J)) THEN
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XX = X(J)
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I=J+1
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100 CONTINUE
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X(I-1) = X(I)
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I = I+1
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IF ((I.LE.N)) then
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if (X(I).LT.XX) GOTO 100
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endif
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X(I-1) = XX
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ENDIF
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ENDDO
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RETURN
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END
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