雪竹的天空

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Robotic Sort

Time limit: 2 Seconds   Memory limit: 32768K  
Total Submit: 12   Accepted Submit: 4  

Somewhere deep in the Czech Technical University buildings, there are laboratories for examining mechanical and electrical properties of various materials. In one of yesterday's presentations, you have seen how was one of the laboratories changed into a new multimedia lab. But there are still others, serving to their original purposes.

In this task, you are to write software for a robot that handles samples in such a laboratory. Imagine there are material samples lined up on a running belt. The samples have different heights, which may cause troubles to the next processing unit. To eliminate such troubles, we need to sort the samples by their height into the ascending order.

Reordering is done by a mechanical robot arm, which is able to pick up any number of consecutive samples and turn them round, such that their mutual order is reversed. In other words, one robot operation can reverse the order of samples on positions between A and B.

A possible way to sort the samples is to find the position of the smallest one (P1) and reverse the order between positions 1 and P1, which causes the smallest sample to become first. Then we find the second one on position P2 and reverse the order between 2 and P2. Then the third sample is located etc.

 

 

The picture shows a simple example of 6 samples. The smallest one is on the 4th position, therefore, the robot arm reverses the first 4 samples. The second smallest sample is the last one, so the next robot operation will reverse the order of five samples on positions 2 - 6. The third step will be to reverse the samples 3 - 4 etc.

Your task is to find the correct sequence of reversal operations that will sort the samples using the above algorithm. If there are more samples with the same height, their mutual order must be preserved: the one that was given first in the initial order must be placed before the others in the final order too.

Input

The input consists of several scenarios. Each scenario is described by two lines. The first line contains one integer number N, the number of samples, 1 <= N <= 100 000. The second line lists exactly N space-separated positive integers, they specify the heights of individual samples and their initial order.

The last scenario is followed by a line containing zero.

Output

For each scenario, output one line with exactly N integers P1, P2, . . .PN, separated by a space. Each Pi must be an integer (1 <= Pi <= N) giving the position of the i-th sample just before the i-th reversal operation.

Note that if a sample is already on its correct position Pi, you should output the number Pi anyway, indicating that the "interval between Pi and Pi" (a single sample) should be reversed.

Sample Input

6
3 4 5 1 6 2
4
3 3 2 1
0

Sample Output

4 6 4 5 6 6
4 2 4 4

 


Problem Source: The 2007 ACM Central Europe Programming Contest

 

 


  1/*
  2* Sample solution to the Sort problem.
  3* Central Europe Regional Contest 2007.
  4*
  5* Zdenek Dvorak, 2007
  6*/

  7
  8#include <stdio.h>
  9#include <stdlib.h>
 10
 11#define MAXN 200000
 12static int n, input[MAXN], ipermut[MAXN], permut[MAXN];
 13
 14struct node
 15{
 16int reverse, size, present;
 17struct node *l, *r, *f;
 18}
;
 19
 20static struct node path_nodes[MAXN];
 21static void
 22dump_tree (struct node *t, int reverse)
 23{
 24struct node *l, *r;
 25
 26reverse ^= t->reverse;
 27
 28= reverse ? t->r : t->l;
 29= reverse ? t->l : t->r;
 30
 31if (l)
 32    dump_tree (l, reverse);
 33
 34if (t->present)
 35    printf ("%d ", (int) (t - path_nodes));
 36
 37if (r)
 38    dump_tree (r, reverse);
 39}

 40
 41static void
 42dump_path (void)
 43{
 44int i;
 45
 46printf ("\n");
 47for (i = 0; i < n; i++)
 48    if (!path_nodes[i].f)
 49      {
 50printf ("--- ");
 51dump_tree (&path_nodes[i], 0);
 52      }

 53
 54printf (" ---\n");
 55}

 56
 57static void
 58clean_reverse (struct node *tree)
 59{
 60struct node *t;
 61
 62if (!tree->reverse)
 63    return;
 64
 65= tree->l; tree->= tree->r; tree->= t;
 66if (tree->l)
 67    tree->l->reverse = !tree->l->reverse;
 68if (tree->r)
 69    tree->r->reverse = !tree->r->reverse;
 70tree->reverse = 0;
 71}

 72
 73static void
 74fix_size (struct node *t)
 75{
 76int lsize = t->? t->l->size : 0;
 77int rsize = t->? t->r->size : 0;
 78
 79t->size = lsize + rsize + t->present;
 80}

 81
 82static void
 83rotate (struct node *t, struct node *f)
 84{
 85struct node **ftlink, **alink, *a;
 86
 87if (f->== t)
 88    {
 89      ftlink = &f->r;
 90      alink = &t->l;
 91    }

 92else
 93    {
 94      ftlink = &f->l;
 95      alink = &t->r;
 96    }

 97
 98t->= f->f;
 99if (f->f)
100    {
101      if (f->f->== f)
102f->f->= t;
103      else
104f->f->= t;
105    }

106
107= *alink;
108*ftlink = a;
109if (a)
110    a->= f;
111
112*alink = f;
113f->= t;
114
115fix_size (f);
116fix_size (t);
117}

118
119static void
120zigzig (struct node *t, struct node *f, struct node *gf)
121{
122struct node **gfflink, **ftlink, **blink, **clink, *b, *c;
123if (gf->== f)
124    {
125      gfflink = &gf->r;
126      ftlink = &f->r;
127      blink = &f->l;
128      clink = &t->l;
129    }

130else
131    {
132      gfflink = &gf->l;
133      ftlink = &f->l;
134      blink = &f->r;
135      clink = &t->r;
136    }

137
138= *blink;
139= *clink;
140
141t->= gf->f;
142if (gf->f)
143    {
144      if (gf->f->== gf)
145gf->f->= t;
146      else
147gf->f->= t;
148    }

149
150*clink = f;
151f->= t;
152
153*blink = gf;
154gf->= f;
155
156*gfflink = b;
157if (b)
158    b->= gf;
159
160*ftlink = c;
161if (c)
162    c->= f;
163
164fix_size (gf);
165fix_size (f);
166fix_size (t);
167}

168
169static void
170zigzag (struct node *t, struct node *f, struct node *gf)
171{
172struct node **gfflink, **ftlink, **blink, **clink, *b, *c;
173if (gf->== f)
174    {
175      gfflink = &gf->l;
176      ftlink = &f->r;
177      blink = &t->l;
178      clink = &t->r;
179    }

180else
181    {
182      gfflink = &gf->r;
183      ftlink = &f->l;
184      blink = &t->r;
185      clink = &t->l;
186    }

187
188= *blink;
189= *clink;
190
191t->= gf->f;
192if (gf->f)
193    {
194      if (gf->f->== gf)
195gf->f->= t;
196      else
197gf->f->= t;
198    }

199
200*clink = gf;
201gf->= t;
202
203*blink = f;
204f->= t;
205
206*gfflink = c;
207if (c)
208    c->= gf;
209
210*ftlink = b;
211if (b)
212    b->= f;
213
214fix_size (gf);
215fix_size (f);
216fix_size (t);
217}

218
219static void
220double_rotate (struct node *t, struct node *f, struct node *gf)
221{
222if ((gf->== f) == (f->== t))
223    zigzig (t, f, gf);
224else
225    zigzag (t, f, gf);
226}

227
228static void
229splay (struct node *t)
230{
231struct node *f, *gf;
232
233while (t->f)
234    {
235      f = t->f;
236
237      gf = f->f;
238      if (!gf)
239{
240   clean_reverse (f);
241   clean_reverse (t);
242   rotate (t, f);
243   return;
244}

245      clean_reverse (gf);
246      clean_reverse (f);
247      clean_reverse (t);
248
249      double_rotate (t, f, gf);
250    }

251
252clean_reverse (t);
253}

254
255static int
256rotfirst_and_delete (struct node *t)
257{
258int pos;
259
260splay (t);
261/*printf ("After splay: ");
262dump_path();
263printf ("\n");*/

264if (t->l)
265    {
266      t->l->reverse = !t->l->reverse;
267      pos = t->l->size;
268    }

269else
270    pos = 0;
271
272t->present = 0;
273t->size--;
274
275return pos;
276}

277
278static int
279cmp_input (const void *a, const void *b)
280{
281int va = *(int *) a;
282int vb = *(int *) b;
283
284if (input[va] != input[vb])
285    return input[va] - input[vb];
286
287return va - vb;
288}

289
290int
291main (void)
292{
293int i;
294
295while (1)
296    {
297      scanf ("%d"&n);
298      if (!n)
299return 0;
300
301      for (i = 0; i < n; i++)
302scanf ("%d"&input[i]);
303
304      for (i = 0; i < n; i++)
305ipermut[i] = i;
306
307      qsort (ipermut, n, sizeof (int), cmp_input);
308
309      for (i = 0; i < n; i++)
310permut[ipermut[i]] = i;
311
312/*      for (i = 0; i < n; i++)
313printf ("%d ", permut[i]);
314      printf ("\n");*/

315      for (i = 0; i < n; i++)
316{
317   path_nodes[permut[i]].reverse = 0;
318   path_nodes[permut[i]].size = i + 1;
319   path_nodes[permut[i]].present = 1;
320   path_nodes[permut[i]].l = i > 0 ? &path_nodes[permut[i-1]] : NULL;
321   path_nodes[permut[i]].r = NULL;
322   path_nodes[permut[i]].f = i < n - 1 ? &path_nodes[permut[i+1]] : NULL;
323}

324
325      for (i = 0; i < n; i++)
326{
327   /*dump_path ();*/
328   printf ("%d%c", rotfirst_and_delete (&path_nodes[i]) + i + 1, i+1 < n ? ' ' : '\n');
329}

330    }

331}

332
posted on 2008-08-31 00:49 雪竹的天空( theorix ) 阅读(510) 评论(0)  编辑 收藏 引用

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