hdu 2609 How many

最新推荐文章于 2024-11-10 21:49:48 发布

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#KMP算法 #set

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本文介绍了一种算法，用于解决给定多个等长字符串（代表项链），通过旋转判断其等价性，并统计不同种类的项链数量的问题。该算法首先确定最小等价串，然后使用 set 进行去重并统计不同种类的数量。

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How many

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3251 Accepted Submission(s): 1446

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

Sample Output

1
2

题不是很难，怪我没想到。借口，借口，还是能力不够。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
char str[10005][105];
char st[10005][105];
int mi_ma(char *s,int len)
{
    int i = 0, j = 1, k = 0;
    while(i < len && j < len && k < len)
    {
        int t = s[(j + k) % len] - s[(i + k) % len];
        if(t == 0)
            k++;
        else
        {
            if(t > 0)
                j += k + 1;
            else
                i += k + 1;
            if(i == j)
                j++;
            k = 0;
        }
    }
    return min(i, j);
}
int main()
{
    int n;
    while(cin >> n)
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%s", str[i]);
            int len=strlen(str[i]);
            int t = mi_ma(str[i],len);
            int j=0,k;
            for(k=t;k<len;k++)
            	st[i][j++]=str[i][k];
            for(k=0;k<t;k++)
            	st[i][j++]=str[i][k];
            st[i][j]='\0';
        }
        set<string>ff;
        for(int i=0;i<n;i++)
        	ff.insert(st[i]);
        cout<<ff.size()<<endl;
    }
    return 0;
}

参考博客

http://blog.youkuaiyun.com/piaocoder/article/details/48447193