hdu 4712 Hamming Distance(随机算法，4级)

Hamming Distance

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 892    Accepted Submission(s): 322

Problem Description

(From wikipedia) For binary strings a and b the Hamming distance is equal to the number of ones in a XOR b. For calculating Hamming distance between two strings a and b, they must have equal length.
Now given N different binary strings, please calculate the minimum Hamming distance between every pair of strings.

 

Input

The first line of the input is an integer T, the number of test cases.(0<T<=20) Then T test case followed. The first line of each test case is an integer N (2<=N<=100000), the number of different binary strings. Then N lines followed, each of the next N line is a string consist of five characters. Each character is '0'-'9' or 'A'-'F', it represents the hexadecimal code of the binary string. For example, the hexadecimal code "12345" represents binary string "00010010001101000101".

 

Output

For each test case, output the minimum Hamming distance between every pair of strings.

 

Sample Input

   
   
   
   

    
    
    
    2
2
12345
54321
4
12345
6789A
BCDEF
0137F
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    6
7
   
   
   
   

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

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思路：随机算法

#include<cstdio>
#include<iostream>
#include<cstring>
#include<ctime>
#include<cstdlib>
using namespace std;
const int mm=1e5+9;
int f[mm];
int get(int x)
{ int ret=0;
  while(x)
  {
    ++ret;x=x&(x-1);
  }
  return ret;
}
int main()
{
  int cas,n,test;
  while(~scanf("%d",&cas))
  while(cas--)
  {
    scanf("%d",&n);
    for(int i=0;i<n;++i)
      scanf("%x",&f[i]);
    test=mm;
    srand((int)time(0));
    int a,b,ans=999;
    while(test--)
    {
      a=rand()%n;
      b=rand()%n;
      if(a==b)continue;
      ans=min(ans,get(f[a]^f[b]));
    }
    printf("%d\n",ans);
  }
}