• hdu4712 Hamming Distance


    Hamming Distance

    Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
    Total Submission(s): 1051 Accepted Submission(s): 396

    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
     
    Recommend
    liuyiding
    用的预处理,随机算法,还是快了很多,主要用的原理也就是a^b=c,则c^b=a;这一点,就可以预处理,得出结果,我们可以看出最坏的复杂度,就是n*2^20,从理论上说,也不比n*n的复杂度,要好到哪里,但是,我们可以发现如果,n很大,那么结果就一定很小,这样,就可以得到最优结果后就退出了,别的什么bfs dfs,也是这种算法的变形,其次我们这题可以用随算法,枚举10w次就可以了,但个人认为不太好!
    #include <iostream>
    #include <stdio.h>
    #include <vector>
    #include <string.h>
    using namespace std;
    #define MAXN 100050
    vector<int > vec[23];
    int hash[1<<21],pri[MAXN];
    char str[]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    void init()
    {
        int i;
        for(i=0;i<=21;i++)
        vec[i].clear();
        int ii=1<<21;
        for(i=0;i<ii;i++)
        {
            int ans=0;
            for(int j=0;j<21;j++)
            {
                if(i&(1<<j))ans++;
            }
            vec[ans].push_back(i);
        }
    }
    int main()
    {
        init();
        int tcase,i,j,k,n;
        char str[20],c;
        scanf("%d",&tcase);
        while(tcase--)
        {
            memset(hash,0,sizeof(hash));
            scanf("%d",&n);
            bool flag=true;
            for(i=0;i<n;i++)
            {
                int ans=0;
                scanf("%X",&pri[i]);
               // printf("%d f
    ",pri[i]);
                hash[pri[i]]++;
                if(flag&&hash[pri[i]]>=2)
                flag=false;
    
            }
            if(!flag)
            {
                printf("0
    ");
                continue;
            }
            for(i=1;i<=20;i++)
            {
                for(j=0;j<n;j++)
                for(k=0;k<vec[i].size();k++)
                {
                    if(hash[pri[j]^vec[i][k]])
                    {
                        printf("%d
    ",i);
                        goto my;
                    }
                }
            }
            my:;
    
        }
        return 0;
    }
    


  • 相关阅读:
    Spring第一次测试错题解析
    正则回顾
    Spring经典---AOP
    动态代理
    MyBatis第一次测试卷---错题分析
    JS中对数组元素进行增删改移
    限制条件补全代码系列题
    字符串去空格
    数组去重
    数组排序
  • 原文地址:https://www.cnblogs.com/pangblog/p/3315483.html
Copyright © 2020-2023  润新知