(字典树)How many--hdu--2609

http://acm.hdu.edu.cn/showproblem.php?pid=2609

 

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1707    Accepted Submission(s): 693

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

4

0110

1100

1001

0011

4

1010

0101

 

1000

0001

 

 

Sample Output

1

2

 

 

用到了上一题的查找最小字典串的方法

一看题没思路了，看看学长是用字典树写的，还从没有用字典树写过题，勉强写了，不过这次可以好好体会体会字典树，但既然是在KMP专题里出现的，想必应该也可以用KMP来写吧，再搜搜

 

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

const int maxn = 205;

char s1[maxn], s[maxn], a[maxn];

struct node
{
    struct node *next[2];
};

int GetMin(char S[], int N)
{
    int i=0, j=1;

    while(i<N && j<N)
    {
        int k = 0;

        while(S[i+k]==S[j+k] && k<N)
            k++;

        if(k==N)
            break;

        if(S[i+k]<S[j+k])
        {
            if(j+k>i)
                j += k+1;
            else
                j = i+1;
        }
        else
        {
            if(i+k>j)
                i += k+1;
            else
                i = j+1;
        }
    }

    return min(i, j);
}

int FindTree(node *head, char s[])
{
     node *p=head;
     int i, flag = 0;

     for(i=0; s[i]; i++)
     {
        int k = s[i]-'0';

        if(p->next[k]==NULL)
        {
            flag = 1;
            p->next[k] = new node();
        }

        p = p->next[k];
     }

     return flag;
}

void Free(node *head)
{
    node *p = head;
    for(int i=0; i<2; i++)
        if(p->next[i]!=NULL)
        Free(p->next[i]);
    free(p);
}


int main()
{
    int n;

    while(scanf("%d", &n)!=EOF)
    {
        int sum = 0;
        node *head = new node();

        while(n--)
        {
            scanf("%s", s1);
            int len = strlen(s1);

            strcpy(a, s1);
            strcat(a, s1);

            int index = GetMin(a, len);
            strncpy(s, a+index, len);

            int ans = FindTree(head, s);

            if(ans)  sum ++;
        }

        printf("%d\n", sum);

        Free(head);
    }
    return 0;
}