Snowflake Snow Snowflakes
Time Limit: 4000MS   Memory Limit: 65536K
Total Submissions: 35642   Accepted: 9373

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.

题意是有很多片雪花,每一个雪花有六片,问这些雪花中是否有相同的雪花,注意雪花可以旋转,即可以循环向前移动或是向后移动,变成相等的情况。

两两比对会TLE,哈希做,六个值的和mod一个质数 作为key,然后有 和相等 的情况下,逐一比对是否含有完全相等的情况。

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <map>
#pragma warning(disable:4996)
using namespace std; const int N1 = 15000;
const int N2 = 100;
int num,m[15000]; struct no {
int sno[7];
}node[N1][N2]; bool check(no node_one,int pos, int x)
{
int i, j;
bool flag = false;
for (j = 0; j < 6; j++)
{
for (i = 1; i <= 6; i++)
{
if (node_one.sno[i] != node[pos][x].sno[(i + j) % 6 + 1])
{
break;
}
if (i == 6)
{
return true;
}
}
} for (j = 7; j >= 2 ; j--)
{
for (i = 1; i <= 6; i++)
{
if (node_one.sno[i] != node[pos][x].sno[(j-i+6)%6+1])
{
break;
}
if (i == 6)
return true;
}
}
return flag;
} int main()
{ int i,j,pos;
bool flag = true;
no node_one; memset(m, 0, sizeof(m));
scanf("%d", &num); for (i = 1; i <= num; i++)
{
for (j = 1; j <= 6; j++)
{
scanf("%d",&node_one.sno[j]);
}
pos = (node_one.sno[1] + node_one.sno[2] + node_one.sno[3] + node_one.sno[4] + node_one.sno[5] + node_one.sno[6])%14997; for (j = 0; j < m[pos]; j++)
{
if (check(node_one,pos,j))
{
puts("Twin snowflakes found.");
return 0;
}
}
node[pos][m[pos]] = node_one;
m[pos]++;
}
puts("No two snowflakes are alike.");
return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

POJ 3349:Snowflake Snow Snowflakes 六片雪花找相同的 哈希的更多相关文章

  1. [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 ...

  2. 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 ...

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

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

  4. 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 ...

  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

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

  7. POJ 3349 Snowflake Snow Snowflakes(哈希表)

    题意:判断有没有两朵相同的雪花.每朵雪花有六瓣,比较花瓣长度的方法看是否是一样的,如果对应的arms有相同的长度说明是一样的.给出n朵,只要有两朵是一样的就输出有Twin snowflakes fou ...

  8. POJ 3349 Snowflake Snow Snowflakes (哈希表)

    题意:每片雪花有六瓣,给出n片雪花,六瓣花瓣的长度按顺时针或逆时针给出,判断其中有没有相同的雪花(六瓣花瓣的长度相同) 思路:如果直接遍历会超时,我试过.这里要用哈希表,哈希表的关键码key用六瓣花瓣 ...

  9. POJ - 3349 Snowflake Snow Snowflakes (哈希)

    题意:给定n(0 < n ≤ 100000)个雪花,每个雪花有6个花瓣(花瓣具有一定的长度),问是否存在两个相同的雪花.若两个雪花以某个花瓣为起点顺时针或逆时针各花瓣长度依次相同,则认为两花瓣相 ...

随机推荐

  1. python--脚本传参与shell脚本传参(位置参数)

    写一个最简单的shell脚本,了解shell脚本是如何传参 1. vim test1.sh name=$1 age=$2 echo ${name} echo ${age} 2.调用脚本并传参 sh t ...

  2. leetCode练题——13. Roman to Integer

    1.题目13. Roman to Integer Roman numerals are represented by seven different symbols: I, V, X, L, C, D ...

  3. GCC 升级

    1.下载源码 wget ftp://ftp.gnu.org/gnu/gcc/gcc-4.8.5/gcc-4.8.5.tar.gz 2.下载依赖包编译安装 GCC 需要依赖 mpc,mpfr,gmp包. ...

  4. vmware fusion nat网络模式设置固定ip

    最近想在本地用虚拟环境搭一个k8s环境,但是发现虚拟机的ip会不定时自动变化,导致mosh客户端连接经常中断.于是就想让虚拟机的ip固定住,不再变动. mac 上的 vmware fusion 设置固 ...

  5. day4-2数组及方法

    数组: Js数组 可以存放任意数据类型的数据 如果索引大于数组的长度,数组自动增加到该索引值加1的长度 var arr = ["terry","larry",& ...

  6. Java开发神器Lombok的使用与原理

    在面向对象编程中必不可少需要在代码中定义对象模型,而在基于Java的业务平台开发实践中尤其如此.相信大家在平时开发中也深有感触,本来是没有多少代码开发量的,但是因为定义的业务模型对象比较多,而需要重复 ...

  7. 如何优雅地根治null值引起的Bug!

    本人免费整理了Java高级资料,涵盖了Java.Redis.MongoDB.MySQL.Zookeeper.Spring Cloud.Dubbo高并发分布式等教程,一共30G,需要自己领取.传送门:h ...

  8. leetcode菜鸡斗智斗勇系列(5)--- 寻找拥有偶数数位的数字

    1.原题: https://leetcode.com/problems/find-numbers-with-even-number-of-digits/ Given an array nums of ...

  9. Node.js npm基础安装配置&创建第一个VUE项目

    使用之前,我们先来明白这几个东西是用来干什么的. node.js: 一种javascript的运行环境,能够使得javascript脱离浏览器运行.Node.js的出现,使得前后端使用同一种语言,统一 ...

  10. 解决Missing artifact com.microsoft.sqlserver:sqljdbc4:jar:4.0问题

    当我们项目中用到的数据库为sql server时  我们一般在maven项目的pom.xml只添加依赖: <dependency>    <groupId>com.micros ...