TOJ 3771 HDU 4162 Shape Number / 最小表示法

Shape Number

时间限制(普通/Java):10000MS/30000MS 运行内存限制:65536KByte



描述

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.

0011002620201167612201100262020116761220
11002620201167612200
...
20011002620201167612

In this case, 00110026202011676122 is the shape number of the shape above.

输入

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.

输出

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

样例输入

22234446466001207560
12075602223444646600

样例输出

0011002620201167612200110026202011676122

就是首先按题意生成一个字符串 这个字符串是循环的 求它字典序最小的一个 可以学习一下最小表示法

#include <stdio.h>
#include <string>
char a[300010],str[300010];
int work(int m)
{
    int i,j,l;
    i=0; j=1;
    while(i<m && j<m)
    {
        for(l=0;l<m;l++) 
            if(str[(i+l)%m]!=str[(j+l)%m]) break;
        if(l>m) break;
        if(str[(i+l)%m] > str[(j+l)%m])
            i=i+l+1;
        else
            j=j+l+1;
        if(i==j) j=i+1;
    }
    if(i<j) return i;
    return j;
}
main()
{
	int i,len,k;
	while(scanf("%s",a)!=EOF)
	{
		len = strlen(a);
		for(i = 0;i < len; i++)
		{
			if(a[(i+1)%len]>=a[i])
				str[i] = a[(i+1)%len] - a[i] + '0';
			else
				str[i] = 8 + a[(i+1)%len] - a[i] + '0';
		}
		k = work(len);
		//printf("%d\n",k);
		for(i = 0; i < len; i++)
		{
			printf("%c",str[(i+k)%len]);
		}
		puts("");
	}
}