悲催的科学匠人 - 冷水's blog

[zz]FORTRAN版非递归快速排序

在二叉树离散网格单元时,最初使用递归的,但是网格单元到百万量级时,排序程序就出错了。怀疑是递归耗尽系统堆栈的问题,搜到一个非递归版本:  http://www.fortran.com/fortran/quick_sort2.f   百万量级也不是问题了。

 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
      PROGRAM test_nrqsort
      REAL,ALLOCATABLE :: array(:)
      INTEGER,ALLOCATABLE :: idx(:)
      INTEGER N,i
      INTEGER,PARAMETER :: seed = 86456
       
      !READ(*,*) N
      N = 5000000
      ALLOCATE(array(N),idx(N))
      CALL srand(seed)
      DO i=1,N;
         array(i) = rand()
      ENDDO
      WRITE(*,*) 'sorting'
      CALL SORTRX(N,array,idx)
      !DO i=1,N
      !  WRITE(*,*) array(idx(i))
      !ENDDO
      DO i=2,N
      IF(array(idx(i)) .lt. array(idx(i-1)) ) THEN
        WRITE(*,*) 'error', array(idx(i)), array(idx(i-1))
      ENDIF
      ENDDO
      DEALLOCATE(array,idx)
      END
 
C From Leonard J. Moss of SLAC:
 
C Here's a hybrid QuickSort I wrote a number of years ago.  It's
C based on suggestions in Knuth, Volume 3, and performs much better
C than a pure QuickSort on short or partially ordered input arrays. 
 
      SUBROUTINE SORTRX(N,DATA,INDEX)
C===================================================================
C
C     SORTRX -- SORT, Real input, indeX output
C
C
C     Input:  N     INTEGER
C             DATA  REAL
C
C     Output: INDEX INTEGER (DIMENSION N)
C
C This routine performs an in-memory sort of the first N elements of
C array DATA, returning into array INDEX the indices of elements of
C DATA arranged in ascending order.  Thus,
C
C    DATA(INDEX(1)) will be the smallest number in array DATA;
C    DATA(INDEX(N)) will be the largest number in DATA.
C
C The original data is not physically rearranged.  The original order
C of equal input values is not necessarily preserved.
C
C===================================================================
C
C SORTRX uses a hybrid QuickSort algorithm, based on several
C suggestions in Knuth, Volume 3, Section 5.2.2.  In particular, the
C "pivot key" [my term] for dividing each subsequence is chosen to be
C the median of the first, last, and middle values of the subsequence;
C and the QuickSort is cut off when a subsequence has 9 or fewer
C elements, and a straight insertion sort of the entire array is done
C at the end.  The result is comparable to a pure insertion sort for
C very short arrays, and very fast for very large arrays (of order 12
C micro-sec/element on the 3081K for arrays of 10K elements).  It is
C also not subject to the poor performance of the pure QuickSort on
C partially ordered data.
C
C Created:  15 Jul 1986  Len Moss
C
C===================================================================
  
      INTEGER   N,INDEX(N)
      REAL      DATA(N)
  
      INTEGER   LSTK(31),RSTK(31),ISTK
      INTEGER   L,R,I,J,P,INDEXP,INDEXT
      REAL      DATAP
  
C     QuickSort Cutoff
C
C     Quit QuickSort-ing when a subsequence contains M or fewer
C     elements and finish off at end with straight insertion sort.
C     According to Knuth, V.3, the optimum value of M is around 9.
  
      INTEGER   M
      PARAMETER (M=9)
  
C===================================================================
C
C     Make initial guess for INDEX
  
      DO 50 I=1,N
         INDEX(I)=I
   50    CONTINUE
  
C     If array is short, skip QuickSort and go directly to
C     the straight insertion sort.
  
      IF (N.LE.M) GOTO 900
  
C===================================================================
C
C     QuickSort
C
C     The "Qn:"s correspond roughly to steps in Algorithm Q,
C     Knuth, V.3, PP.116-117, modified to select the median
C     of the first, last, and middle elements as the "pivot
C     key" (in Knuth's notation, "K").  Also modified to leave
C     data in place and produce an INDEX array.  To simplify
C     comments, let DATA[I]=DATA(INDEX(I)).
  
C Q1: Initialize
      ISTK=0
      L=1
      R=N
  
  200 CONTINUE
  
C Q2: Sort the subsequence DATA[L]..DATA[R].
C
C     At this point, DATA[l] <= DATA[m] <= DATA[r] for all l < L,
C     r > R, and L <= m <= R.  (First time through, there is no
C     DATA for l < L or r > R.)
  
      I=L
      J=R
  
C Q2.5: Select pivot key
C
C     Let the pivot, P, be the midpoint of this subsequence,
C     P=(L+R)/2; then rearrange INDEX(L), INDEX(P), and INDEX(R)
C     so the corresponding DATA values are in increasing order.
C     The pivot key, DATAP, is then DATA[P].
  
      P=(L+R)/2
      INDEXP=INDEX(P)
      DATAP=DATA(INDEXP)
  
      IF (DATA(INDEX(L)) .GT. DATAP) THEN
         INDEX(P)=INDEX(L)
         INDEX(L)=INDEXP
         INDEXP=INDEX(P)
         DATAP=DATA(INDEXP)
      ENDIF
  
      IF (DATAP .GT. DATA(INDEX(R))) THEN
         IF (DATA(INDEX(L)) .GT. DATA(INDEX(R))) THEN
            INDEX(P)=INDEX(L)
            INDEX(L)=INDEX(R)
         ELSE
            INDEX(P)=INDEX(R)
         ENDIF
         INDEX(R)=INDEXP
         INDEXP=INDEX(P)
         DATAP=DATA(INDEXP)
      ENDIF
  
C     Now we swap values between the right and left sides and/or
C     move DATAP until all smaller values are on the left and all
C     larger values are on the right.  Neither the left or right
C     side will be internally ordered yet; however, DATAP will be
C     in its final position.
  
  300 CONTINUE
  
C Q3: Search for datum on left >= DATAP
C
C     At this point, DATA[L] <= DATAP.  We can therefore start scanning
C     up from L, looking for a value >= DATAP (this scan is guaranteed
C     to terminate since we initially placed DATAP near the middle of
C     the subsequence).
  
         I=I+1
         IF (DATA(INDEX(I)).LT.DATAP) GOTO 300
  
  400 CONTINUE
  
C Q4: Search for datum on right <= DATAP
C
C     At this point, DATA[R] >= DATAP.  We can therefore start scanning
C     down from R, looking for a value <= DATAP (this scan is guaranteed
C     to terminate since we initially placed DATAP near the middle of
C     the subsequence).
  
         J=J-1
         IF (DATA(INDEX(J)).GT.DATAP) GOTO 400
  
C Q5: Have the two scans collided?
  
      IF (I.LT.J) THEN
  
C Q6: No, interchange DATA[I] <--> DATA[J] and continue
  
         INDEXT=INDEX(I)
         INDEX(I)=INDEX(J)
         INDEX(J)=INDEXT
         GOTO 300
      ELSE
  
C Q7: Yes, select next subsequence to sort
C
C     At this point, I >= J and DATA[l] <= DATA[I] == DATAP <= DATA[r],
C     for all L <= l < I and J < r <= R.  If both subsequences are
C     more than M elements long, push the longer one on the stack and
C     go back to QuickSort the shorter; if only one is more than M
C     elements long, go back and QuickSort it; otherwise, pop a
C     subsequence off the stack and QuickSort it.
  
         IF (R-J .GE. I-L .AND. I-L .GT. M) THEN
            ISTK=ISTK+1
            LSTK(ISTK)=J+1
            RSTK(ISTK)=R
            R=I-1
         ELSE IF (I-L .GT. R-J .AND. R-J .GT. M) THEN
            ISTK=ISTK+1
            LSTK(ISTK)=L
            RSTK(ISTK)=I-1
            L=J+1
         ELSE IF (R-J .GT. M) THEN
            L=J+1
         ELSE IF (I-L .GT. M) THEN
            R=I-1
         ELSE
C Q8: Pop the stack, or terminate QuickSort if empty
            IF (ISTK.LT.1) GOTO 900
            L=LSTK(ISTK)
            R=RSTK(ISTK)
            ISTK=ISTK-1
         ENDIF
         GOTO 200
      ENDIF
  
  900 CONTINUE
  
C===================================================================
C
C Q9: Straight Insertion sort
  
      DO 950 I=2,N
         IF (DATA(INDEX(I-1)) .GT. DATA(INDEX(I))) THEN
            INDEXP=INDEX(I)
            DATAP=DATA(INDEXP)
            P=I-1
  920       CONTINUE
               INDEX(P+1) = INDEX(P)
               P=P-1
               IF (P.GT.0) THEN
                  IF (DATA(INDEX(P)).GT.DATAP) GOTO 920
               ENDIF
            INDEX(P+1) = INDEXP
         ENDIF
  950    CONTINUE
  
C===================================================================
C
C     All done
  
      END




Host by is-Programmer.com | Power by Chito 1.3.3 beta | © 2007 LinuxGem | Design by Matthew "Agent Spork" McGee