【HDU 4162】Shape Number(一阶差分链码+最小表示法)

【HDU 4162】Shape Number(一阶差分链码+最小表示法)

Shape Number

Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1249 Accepted Submission(s): 593



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
00110026202011676122

Source

The 2011 Rocky Mountain Regional Contest



Recommend

lcy | We have carefully selected several similar problems for you: 4161 4163 4164 4165 4168



先处理一下输入串 说是啥求一阶差分链码。。反正根据题意就是把str[i]变成str[i+1]转到str[i]需要的步数 逆时针按图转

转换后跑一次最小表示法找出同构字符串中最小字典序即可

最小表示法这有几篇好文不错

http://www.cnblogs.com/wuhenqs/p/3201053.html 这是本题题解 感觉讲最小表示讲的挺详细的

http://blog.csdn.net/zy691357966/article/details/39854359

/article/1819422.html 这篇讲的比较浅显 能很快明白 但不太好深刻理解

此题代码如下:

#include <bits/stdc++.h>

using namespace std;

char str[300100];

int Minimze(char *s, int len)
{
    int i,j,k,t;
    i = k = 0;
    j = 1;

    while(i < len && j < len && k < len)
    {
        t = s[(i+k)%len] - s[(j+k)%len];
        if(!t) ++k;
        else
        {
            if(t > 0) i = i+k+1;
            else j = j+k+1;
            if(i == j) ++j;
            k = 0;
        }
    }
    return min(i,j);
}

int main()
{
    int len;
    while(~scanf("%s",str))
    {
        len = strlen(str);
        str[len] = str[0];
        for(int i = 0; i < len; ++i)
            str[i] = (str[i+1]-str[i]+8)%8+'0';
        str[len] = 0;
        int pos = Minimze(str,len);
        printf("%s",str+pos);
        str[pos] = 0;
        puts(str);
    }
    return 0;
}