hdu 4162 Shape Number

最新推荐文章于 2025-07-14 21:00:27 发布

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最新推荐文章于 2025-07-14 21:00:27 发布

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原文链接：http://www.cnblogs.com/ERKE/p/3832942.html

本文深入探讨了形状编码的概念及其在计算机视觉领域的应用。详细介绍了如何通过计算链码的第一差分来规范化形状编码，以消除旋转和起点选择的影响，最终得到形状的唯一标识符——形状数。文中提供了一个具体的实现算法，包括输入处理、链码规范化和输出形状数的过程，通过实例展示了规范化前后的链码变化。

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Shape Number

Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 798 Accepted Submission(s): 387

Problem Description

In computer vision, a chain code is a sequence of numbers representing directions when following the contour of an object. For example, the following figure shows the contour represented by the chain code 22234446466001207560 (starting at the upper-left corner).

Two chain codes may represent the same shape if the shape has been rotated, or if a different starting point is chosen for the contour. To normalize the code for rotation, we can compute the first difference of the chain code instead. The first difference is obtained by counting the number of direction changes in counterclockwise direction between consecutive elements in the chain code (the last element is consecutive with the first one). In the above code, the first difference is

00110026202011676122
Finally, to normalize for the starting point, we consider all cyclic rotations of the first difference and choose among them the lexicographically smallest such code. The resulting code is called the shape number.
00110026202011676122
01100262020116761220
11002620201167612200
...
20011002620201167612
In this case, 00110026202011676122 is the shape number of the shape above.

Input

The input consists of a number of cases. The input of each case is given in one line, consisting of a chain code of a shape. The length of the chain code is at most 300,000, and all digits in the code are between 0 and 7 inclusive. The contour may intersect itself and needs not trace back to the starting point.

Output

For each case, print the resulting shape number after the normalizations discussed above are performed.

Sample Input

22234446466001207560

12075602223444646600

Sample Output

00110026202011676122

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 #include <string.h>
 5 #include <set>
 6 using namespace std;
 7 char s[400000],a[400000];
 8 int moremin(char s[],int l)
 9 {
10     int i,j,k,t;
11     i=k=0,j=1;
12     while(i<l&&j<l&&k<l)
13     {
14         int t=s[i+k>=l?i+k-l:i+k]-s[j+k>=l?j+k-l:j+k];
15         if(!t)k++;
16         else
17         {
18             if(t>0)i+=k+1;
19             else j+=k+1;
20             if(i==j)j++;
21             k=0;
22         }
23     }
24     return (i>j?j:i);
25 }
26 int main()
27 {
28     int i;
29     while(~scanf("%s",a))
30     {
31         int l=strlen(a);
32         for(i=0;i<l;i++)
33         if(a[i+1>=l?i+1-l:i+1]>=a[i])
34         s[i]=a[i+1>=l?i+1-l:i+1]-a[i]+'0';
35         else s[i]=8+a[i+1>=l?i+1-l:i+1]-a[i]+'0';
36         s[i]='\0';
37         int x=moremin(s,l);
38         for(i=0;i<l;i++)
39         putchar(s[i+x>=l?i+x-l:i+x]);
40         puts("");
41     }
42 }