Snowflake Snow Snowflakes
Time Limit: 4000MS Memory Limit: 65536K
Total Submissions: 27598 Accepted: 7312

Description

You may have heard that no two snowflakes are alike. Your task is to write a program to determine whether this is really true. Your program will read information about a collection of snowflakes, and search for a pair that may be identical. Each snowflake has six arms. For each snowflake, your program will be provided with a measurement of the length of each of the six arms. Any pair of snowflakes which have the same lengths of corresponding arms should be flagged by your program as possibly identical.

Input

The first line of input will contain a single integer n, 0 < n ≤ 100000, the number of snowflakes to follow. This will be followed by n lines, each describing a snowflake. Each snowflake will be described by a line containing six integers (each integer is at least 0 and less than 10000000), the lengths of the arms of the snow ake. The lengths of the arms will be given in order around the snowflake (either clockwise or counterclockwise), but they may begin with any of the six arms. For example, the same snowflake could be described as 1 2 3 4 5 6 or 4 3 2 1 6 5.

Output

If all of the snowflakes are distinct, your program should print the message:
No two snowflakes are alike.
If there is a pair of possibly identical snow akes, your program should print the message:
Twin snowflakes found.

Sample Input

2
1 2 3 4 5 6
4 3 2 1 6 5

Sample Output

Twin snowflakes found.

Source

CCC 2007

hash函数。。。3750ms。。。险过。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

struct ST
{
   int a[6];
};

bool cmp(const ST x,const ST y)
{
    bool flag=true;
    for(int i=0;i<6;i++)
    {
        flag=true;
        for(int j=0;j<6;j++)
        {
            if(x.a[j]!=y.a[(j+i)%6])
            {
                flag=false;
                break;
            }
        }
        if(flag) return true;
    }
    for(int i=0;i<6;i++)
    {
        flag=true;
        for(int j=0;j<6;j++)
        {
            if(x.a[5-j]!=y.a[(j+i)%6])
            {
                flag=false;
                break;
            }
        }
        if(flag) return true;
    }
    return false;
}

int main()
{
    int n;
    ST A; bool flag=false;
    vector<ST> hash[17569];
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        int sum=0;
        for(int j=0;j<6;j++)
        {
            scanf("%d",&A.a[j]);
            sum+=A.a[j];
        }
        hash[sum%17569].push_back(A);
    }
    for(int i=0;i<17569;i++)
    {
        if(flag) continue;
        int nth=hash.size();
        for(int j=0;j<nth;j++)
        {
            for(int k=j+1;k<nth;k++)
            {
                if(cmp(hash[j],hash[k]))
                {
                    puts("Twin snowflakes found.");
                    flag=true;
                    break;
                }
            }
        }
    }
    if(flag==false)
        puts("No two snowflakes are alike.");
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )

POJ 3349 Snowflake Snow Snowflakes的更多相关文章

  1. 哈希—— POJ 3349 Snowflake Snow Snowflakes

    相应POJ题目:点击打开链接 Snowflake Snow Snowflakes Time Limit: 4000MS   Memory Limit: 65536K Total Submissions ...

  2. POJ 3349 Snowflake Snow Snowflakes(简单哈希)

    Snowflake Snow Snowflakes Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 39324   Accep ...

  3. poj 3349:Snowflake Snow Snowflakes(哈希查找,求和取余法+拉链法)

    Snowflake Snow Snowflakes Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 30529   Accep ...

  4. POJ 3349 Snowflake Snow Snowflakes (Hash)

    Snowflake Snow Snowflakes Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 48646   Accep ...

  5. [ACM] POJ 3349 Snowflake Snow Snowflakes(哈希查找,链式解决冲突)

    Snowflake Snow Snowflakes Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 30512   Accep ...

  6. [poj 3349] Snowflake Snow Snowflakes 解题报告 (hash表)

    题目链接:http://poj.org/problem?id=3349 Description You may have heard that no two snowflakes are alike. ...

  7. POJ 3349 Snowflake Snow Snowflakes(哈希)

    http://poj.org/problem?id=3349 题意 :分别给你n片雪花的六个角的长度,让你比较一下这n个雪花有没有相同的. 思路:一开始以为把每一个雪花的六个角的长度sort一下,然后 ...

  8. POJ 3349 Snowflake Snow Snowflakes Hash

    题目链接: http://poj.org/problem?id=3349 #include <stdio.h> #include <string.h> #include < ...

  9. hash应用以及vector的使用简介:POJ 3349 Snowflake Snow Snowflakes

    今天学的hash.说实话还没怎么搞懂,明天有时间把知识点总结写了,今天就小小的写个结题报告吧! 题意: 在n (n<100000)个雪花中判断是否存在两片完全相同的雪花,每片雪花有6个角,每个角 ...

随机推荐

  1. 浩瀚先森(guohao1206.com)

    博客搬家啦,新博客地址:浩瀚先森 http://www.guohao1206.com

  2. javascript 事件传播与事件冒泡,W3C事件模型

    说实话笔者在才工作的时候就听说了什么"事件冒泡",弄了很久才弄个大概,当时理解意思是子级dom元素和父级dom元素都绑定了相同类型的事件,这时如果子级事件触发了父级也会触发,然后这 ...

  3. PHP+MD5

    echo substr(md5("admin"),8,16);//16位MD5加密,16位加密是32位加密从第8个字符开始到16个字符,用substr函数截取了字符得到.. ech ...

  4. Newtonsoft.Json之JArray, JObject, JPropertyJValue

    JObject staff = new JObject(); staff.Add(new JProperty("Name", "Jack")); staff.A ...

  5. php中命名空间的使用

    简单使用 命名空间主要解决函数/类冲突的问题.由于PHP中中不允许函数重载,所以我们要使用的到命名空间的.先看一个简单的例子. <?php namespace A; public functio ...

  6. MySQL配置

    一.登录MySQL 要登录到MySQL只需要使用如下命令. mysql -h localhost -u root -p localhost:IP地址: root:用户名: database:数据库名( ...

  7. 作业一_随笔3_调研Android的开发环境的发展演变

    调研某一移动应用/平台的开发环境的发展演变:Android 其实,一开始,我只知道,苹果手机用IOS系统,其他很多手机时候安卓系统.我百度知道Android开发主要是android studio和Ec ...

  8. .Net身份验证概述

    一直以来,所有的系统基本都会有用户的登陆验证过程,整个过程其实也不难理解,就是对于cookie的解析.微软的.Net平台围绕用户身份验证授权也有好几个版本了,从早期的Membership到Identi ...

  9. 45.Android 第三方开源库收集整理(转)

    原文地址:http://blog.csdn.net/caoyouxing/article/details/42418591 Android开源库 自己一直很喜欢Android开发,就如博客签名一样,  ...

  10. css常用技巧

    去点号 list-style:none; 字体居中 text-align:center; 链接去下划线 text-decoration:none; 鼠标禁止右键 <body oncontextm ...