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.

题意:给n朵雪花,每个雪花都有6条边,找出是否有两个相同的雪花。
思路:n²遍历肯定不行,然后就想到hash,对于存放hash值的邻接表,我用的是vector,事实证明很慢啊,手写前向星快很多。
hash:我用的是直接相加然后取模一个素数的方法,当然还有很多其他方法,各种异或或者骚运算应该都差不多。 (对于equl函数,我错了很久,之前写的是用while找到一个pos位置和b【0】相同,就退出,然后我发现,找到一个不够,也许有下一个不能直接退出)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std; const int SIZE = ;
int n,tot,p=;
int snow[SIZE][];
vector<int>v[SIZE]; int h(int a)
{
int sum=;
for(int i=;i<;i++)
{
sum = (sum + snow[a][i]) % p;
}
return sum % p;
}; bool equl(int a,int b)
{
for(int j=;j<;j++)
{
if(snow[a][j] != snow[b][])continue;
int flag = ;
for(int i=; i<; i++)
{
if(snow[a][(j + i)%] != snow[b][i] )
flag = ;
}
if(flag)
return ;
flag = ;
for(int i=; i<; i++)
{
if(snow[a][(j-i+)%] != snow[b][i])
flag = ;
}
if(flag) return ;
}
return ;
} bool Insert(int a)
{
int val = h(a);
int len = v[val].size();
for(int i=;i<len;i++)
{ if(equl(v[val][i],a))return ;
}
v[val].push_back(a);
return ;
} int main()
{
scanf("%d",&n);
int flag = ;
for(int i=;i<=n;i++)
{
for(int j=;j<;j++)
{
scanf("%d",&snow[i][j]);
}
if(!flag && Insert(i))
{
puts("Twin snowflakes found.");
flag = ;
}
}
if(!flag)
puts("No two snowflakes are alike.");
}

Snowflake Snow Snowflakes POJ - 3349(hash)的更多相关文章

  1. [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. ...

  2. Snowflake Snow Snowflakes POJ - 3349 Hash

    题意:一个雪花有六个角  给出N个雪花 判断有没有相同的(可以随意旋转) 参考:https://blog.csdn.net/alongela/article/details/8245005 注意:参考 ...

  3. Snowflake Snow Snowflakes - poj 3349 (hash函数)

    判断n朵雪花中,是否有完全一样的雪花.简单的hash,将雪花的六个边的权值加起来,记为sum,将sum相等的雪花归为一类,再在这里面根据题意找完全相同的,判断顺时针或者逆时针的所有角是否一模一样. # ...

  4. POJ 3349:Snowflake Snow Snowflakes(数的Hash)

    http://poj.org/problem?id=3349 Snowflake Snow Snowflakes Time Limit: 4000MS   Memory Limit: 65536K T ...

  5. POJ 3349 Snowflake Snow Snowflakes (Hash)

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

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

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

  7. [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 ...

  8. 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 ...

  9. POJ--3349 Snowflake Snow Snowflakes(数字hash)

    链接:Snowflake Snow Snowflakes 判断所有的雪花里面有没有相同的 每次把雪花每个角的值进行相加和相乘 之后hash #include<iostream> #incl ...

随机推荐

  1. python 面向对象编程(高级篇)

    飞机票 面向对象是一种编程方式,此编程方式的实现是基于对 类 和 对象 的使用 类 是一个模板,模板中包装了多个“函数”供使用(可以讲多函数中公用的变量封装到对象中) 对象,根据模板创建的实例(即:对 ...

  2. SQL Server2016导出数据表数据

    SQL Server2016导出数据表数据 高文龙关注0人评论3914人阅读2017-09-22 08:41:56 SQL Server2016导出数据表数据 我们前面已经介绍了很多关于SQL Ser ...

  3. JS和Jquery获取和修改label的值的示例代码

    abel标签在JS和Jquery中使用不能像其他标签一样用value获取它的值,下面有个不错的示例,希望大家可以学习下 来源:  <  JS和Jquery获取和修改label的值的示例代码  & ...

  4. tarjan求lca :并查集+dfs

    //参考博客 https://www.cnblogs.com/jsawz/p/6723221.html#include<bits/stdc++.h> using namespace std ...

  5. node.js 的热更新

    1.安装 npm i supervisor -gd 2.运行 supervisor server.js //server.js 是你自己的服务的js文件

  6. 向github上提交自己的project

    参考博客:https://blog.csdn.net/m0_37725003/article/details/80904824 step I: 创建自己的github账户(username:fourm ...

  7. web和servlet一样的用法但是有区别

  8. PV-UV-QPS

    QPS:每秒查询率(Query Per Second) ,每秒的响应请求数,也即是最大吞吐能力.QPS = req/sec = 请求数/秒QPS统计方式 [一般使用 http_load 进行统计]QP ...

  9. Linux桌面环境安装matlab并创建快捷方式

    安装matlab sudo mkdir -p /mnt/matlab sudo mount -t auto -o loop /home/chris/Downloads/2016b_linux/R201 ...

  10. Linux下查看80端口是否被占用

    方式一: ps -ef |grep 80 方式二: netstat -anp |grep :80 方式三: lsof -i:80 方式四: netstat -tunlp |grep :80 方式五: ...