Hamming Distance

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 797 Accepted Submission(s): 284

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

【题解】：

　　随机函数（没节操）暴力，（网上有大神这么水过的）

　　一开始看到rand随机，就觉得真没节操，后来想了想，这题确实可以这么做，因为字符串长度只有5，所以答案肯定在0-20之间，而随机次数越多得到的答案就是最小值的机会也就越大，所以要尽量在保证不超时的前提下，增加尽量多的随机次数

【code】：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 #include <string.h>
 5 #include <time.h>
 6 
 7 using namespace std;
 8 #define N 100000
 9 
10 char str[N+10][10];
11 int mark[20][20];  //make中存 i^j 的1的个数
12 int arr[]={0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};  //0-F 中1的个数
13 
14 int charToHex(char ch)  //将0-F字符转换成10进制数计算
15 {
16     if(isdigit(ch)) return ch-'0';
17     return ch-'A'+10;
18 }
19 
20 void getMark() //求mark数组
21 {
22     int i,j,s;
23     for(i=0;i<16;i++)
24     {
25         for(j=i;j<16;j++)
26         {
27             s=i^j;
28             mark[i][j]=mark[j][i]=arr[s];
29         }
30     }
31 }
32 
33 int geths(int x,int y) //求x到y的Hamming distance
34 {
35     int i,sum=0;
36     for(i=0;i<5;i++)
37     {
38         int xx = charToHex(str[x][i]);
39         int yy = charToHex(str[y][i]);
40         sum+=mark[xx][yy];
41     }
42     return sum;
43 }
44 
45 int main()
46 {
47     int t;
48     getMark();
49     scanf("%d",&t);
50     while(t--)
51     {
52         int n;
53         scanf("%d",&n);
54         int i;
55         for(i=0;i<n;i++)
56         {
57             scanf("%s",str[i]);
58         }
59         srand(time(NULL));
60         int x,y,mins=100;
61         for(i=0;i<900000;i++) //随机900000次基本能过，在不超时的前提下，随机次数越多越好
62         {
63             x=rand()%n;
64             y=rand()%n;
65             if(x==y)    continue;
66             int temp = geths(x,y);
67             if(mins>temp)   mins=temp;
68         }
69         printf("%d
",mins);
70     }
71     return 0;
72 }
73 /*
74 2
75 2
76 12345
77 54321
78 4
79 12345
80 6789A
81 BCDEF
82 0137F
83 */