hdu 4712 Hamming Distance ( 随机算法混过了 )

Hamming Distance

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 499    Accepted Submission(s): 163

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

 

ps：不知道正解是怎样做，用随机算法水过了。

思路：

答案的范围不大，0~20，数据不强，可以考虑随机算法。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <algorithm>
#define maxn 105
#define maxx 100005
#define INF 0x3f3f3f3f
using namespace std;

int n,m,ans;
int xr[maxn][maxn];
int a[]= {0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};
char s[maxx][6];

int cal(int k1,int k2)
{
    int i,j,t=0,x,y;
    for(i=0; i<5; i++)
    {
        x=s[k1][i];
        if(x>='A'&&x<='F') x=x-'A'+10;
        else x=x-'0';
        y=s[k2][i];
        if(y>='A'&&y<='F') y=y-'A'+10;
        else y=y-'0';
        t+=xr[x][y];
    }
    return t;
}
void solve()
{
    int i,j,t,x,y;
    srand((unsigned)time(NULL));
    for(i=1; i<=100000; i++)
    {
        x=rand()%n;
        y=rand()%n;
        if(x==y) continue ;
        t=cal(x,y);
        if(ans>t) ans=t;
        if(!ans) return ;
    }
}
int main()
{
    int i,j,t;
    for(i=0; i<16; i++)
    {
        for(j=0; j<16; j++)
        {
            xr[i][j]=a[i^j];
        }
    }
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0; i<n; i++)
        {
            scanf("%s",s[i]);
        }
        ans=INF;
        solve();
        printf("%d
",ans);
    }
    return 0;
}