Hamming Distance

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1569    Accepted Submission(s): 616

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   这道题的正常做法是O（n^2）的，很显然这个做法会大大方方的T掉 但是我们可以用随机化，然后悄悄不说话的水过去= =

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <time.h>

using namespace std;
#define N 100000

char str[N+10][10];
int mark[20][20];  //make中存 i^j 的1的个数
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

int charToHex(char ch)  //将0-F字符转换成10进制数计算
{
    if(isdigit(ch)) return ch-'0';
    return ch-'A'+10;
}

void getMark() //求mark数组
{
    int i,j,s;
    for(i=0;i<16;i++)
    {
        for(j=i;j<16;j++)
        {
            s=i^j;
            mark[i][j]=mark[j][i]=arr[s];
        }
    }
}

int geths(int x,int y) //求x到y的Hamming distance
{
    int i,sum=0;
    for(i=0;i<5;i++)
    {
        int xx = charToHex(str[x][i]);
        int yy = charToHex(str[y][i]);
        sum+=mark[xx][yy];
    }
    return sum;
}

int main()
{
    int t;
    getMark();
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int i;
        for(i=0;i<n;i++)
        {
            scanf("%s",str[i]);
        }
        srand(time(NULL));
        int x,y,mins=100;
        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
        {
            x=rand()%n;
            y=rand()%n;
            if(x==y)    continue;
            int temp = geths(x,y);
            if(mins>temp)   mins=temp;
        }
        printf("%d\n",mins);
    }
    return 0;
}
/*
2
2
12345
54321
4
12345
6789A
BCDEF
0137F
*/