[转]浅析“最小表示法”思想在字符串循环同构问题中的应用－HDU2609

ACM偶尔会出现最小表示法的题目，下面是一些文档可以帮助理解。

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How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1117    Accepted Submission(s): 444

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

4011011001001001141010010110000001

Sample Output

12

Author

yifenfei

Source

奋斗的年代

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解释：见上面的pdf。Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor97690422013-12-06 16:16:07Accepted2609109MS2032K866 BC++ 

//第一次用最小表示法#include<iostream>#include<set>#include<string>#include<math.h>using namespace std;int GetMin(const string& src){int i=0,j=1,k=0,len=src.size(),pos;while (i < len && j < len && k < len){pos = src[(i+k)%len]-src[(j+k)%len];if( pos == 0 )++k;else{if( pos > 0 )i = i + k + 1;elsej = j + k + 1;if( i == j )++j;k = 0;}}return i < j ?  i : j;}int main(){int T,pos,diff;string src,strTemp;set<string> Set;while(cin >> T){Set.clear();diff = 0;while( T-- ){cin >> strTemp;pos = GetMin(strTemp);src.assign(strTemp,pos,strTemp.size()-pos);if( pos )src.append(strTemp,0,pos);if( Set.count(src) == 0 ){++diff;Set.insert(src);}}cout << diff << endl;}return 0;}