【最小表示法+set去重】P - How many HDU - 2609

最新推荐文章于 2022-04-13 15:22:58 发布

原创最新推荐文章于 2022-04-13 15:22:58 发布 · 198 阅读

0 ·

CC 4.0 BY-SA版权

字符串同时被 2 个专栏收录

32 篇文章

订阅专栏

STL

7 篇文章

订阅专栏

本文介绍了一种用于计算不同种类项链数量的算法。通过旋转项链并比较其字符序列来判断两个项链是否相同。使用C++实现，该算法适用于长度不超过100的项链，并能够处理最多10000条项链的输入。

How many

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4011 Accepted Submission(s): 1821

Problem Description

Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me
How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some).
For example 0110 express a necklace, you can rotate it. 0110 -> 1100 -> 1001 -> 0011->0110.

Input

The input contains multiple test cases.
Each test case include: first one integers n. (2<=n<=10000)
Next n lines follow. Each line has a equal length character string. (string only include '0','1').

Output

For each test case output a integer , how many different necklaces.

Sample Input

0110

1100

1001

0011

1010

0101

1000

0001

Sample Output

Author

yifenfei

Source

奋斗的年代

#include <bits/stdc++.h>
using namespace std;

int len;
int cal_Min(char ch[])
{
	int i = 0, j = 1, k = 0;
	while (i < len && j < len && k < len)
	{
		char a = ch[(i + k) % len];
		char b = ch[(j + k) % len];

		if (a == b)
			k++;
		else
		{
			if (a > b)
				i = i + k + 1;
			else
				j = j + k + 1;
			k = 0;

			if (i == j)
				j++;
		}
	}
	return min(i, j);
}

set<string> Set;

int main()
{
#ifndef ONLINE_JUDGE
	freopen("C:\\in.txt", "r", stdin);
#endif // ONLINE_JUDGE

	int n;
	while (~scanf("%d", &n))
	{
		Set.clear();
		while (n--)
		{
			char ch[110];
			scanf("%s", ch);

			len = strlen(ch);
			int pos = cal_Min(ch);
			/// 旋转后相同的字符串 = 他们经过不同次旋转得到的最小表示相同

			char t[110];
			int w = 0;
			for (int i = pos; ; i = (i + 1) % len)
			{
				t[w++] = ch[i];
				if (w == len)
					break;
			}
			t[len] = '\0';
			Set.insert(t);
			/// 字符串 set 声明时用string, 插入时用char[]
		}
		printf("%d\n", Set.size());
	}
}