HDU 4712：Hamming Distance

HDU 4712：Hamming Distance

Hamming Distance

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1845    Accepted Submission(s): 740

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

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

迷失在幽谷中的鸟儿，独自飞翔在这偌大的天地间，却不知自己该飞往何方……

题意：给出n个字符串，求每两个串相应位异或值的二进制中1的个数和的最小值！例如： 12345 54321其相应位异或之后为：46064它的每一位二进制和为6

一般的枚举会超时，因为我试过……，然后为了优化，给两个数字异或之后的值打表！然后，然后还是超时……

既然不能遍历每一种情况，那就随机吧！只要控制随机总次数，就可以控制时间长度，至于AC，那就靠运气啦！不过这样的随机总次数，一般都会AC……

#include #include #include #include using namespace std;int cmp[16][16]= {0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,0,2,1,2,1,3,2,2,1,3,2,3,2,4,3,1,2,0,1,2,3,1,2,2,3,1,2,3,4,2,3,2,1,1,0,3,2,2,1,3,2,2,1,4,3,3,2,1,2,2,3,0,1,1,2,2,3,3,4,1,2,2,3,2,1,3,2,1,0,2,1,3,2,4,3,2,1,3,2,2,3,1,2,1,2,0,1,3,4,2,3,2,3,1,2,3,2,2,1,2,1,1,0,4,3,3,2,3,2,2,1,1,2,2,3,2,3,3,4,0,1,1,2,1,2,2,3,2,1,3,2,3,2,4,3,1,0,2,1,2,1,3,2,2,3,1,2,3,4,2,3,1,2,0,1,2,3,1,2,3,2,2,1,4,3,3,2,2,1,1,0,3,2,2,1,2,3,3,4,1,2,2,3,1,2,2,3,0,1,1,2,3,2,4,3,2,1,3,2,2,1,3,2,1,0,2,1,3,4,2,3,2,3,1,2,2,3,1,2,1,2,0,1,4,3,3,2,3,2,2,1,3,2,2,1,2,1,1,0};;char p[1000005][10];int ddd(int q,int w){ int a,b,s=0; for(int i=0; i<5; i++) { char x = p[q][i]; char y = p[w][i]; if(isdigit(x))a=x-'0'; else a=x-'A'+10; if(isdigit(y))b=y-'0'; else b=y-'A'+10; s += cmp[a][b]; } return s;}int main(){ int T; cin>>T; while(T--) { int N; cin>>N; int minn=0xfffffff; for(int i=0; i>p[i]; srand((unsigned)time(NULL)); for(int i=1; i<=100000; i++) { int a=rand()%N; int b=rand()%N; if(a==b)continue; int tmp=ddd(a,b); minn=tmp>minn?minn:tmp; } cout<

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