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。

#include<vector>
#include<stdio.h>
int MOD=;
std::vector<int> snow[];
int arm[][];
int isSame(int s1,int s2){
for(int i=;i<;i++){
int flag=;
for(int j=,k=i;j<;j++,k=(k+)%){
if(arm[s1][j]!=arm[s2][k]){
flag=;
break;
}
}
if(flag) return ;
}
for(int i=;i<;i++){
int flag=;
for(int j=,k=i;j>=;j--,k=(k+)%){
if(arm[s1][j]!=arm[s2][k]){
flag=;
break;
}
}
if(flag) return ;
}
return ;
}
int main() {
int N;
scanf("%d",&N);
int flag=;
for(int i=;i<N;i++){
int sum=;
for(int j=;j<;j++){
scanf("%d",&arm[i][j]);
sum+=arm[i][j];
}
if(flag)
continue;
sum=sum%MOD;
for(int j=;j<snow[sum].size();j++){
flag=isSame(snow[sum][j],i);
if(flag)
break;
}
snow[sum].push_back(i);
}
if(flag)
printf("Twin snowflakes found.\n");
else
printf("No two snowflakes are alike.\n");
return ;
}

附:

 
Time Limit: 4000MS   Memory Limit: 65536K
Total Submissions: 34732   Accepted: 9119

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.

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

  1. 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 ...

  2. Snowflake Snow Snowflakes POJ - 3349(hash)

    You may have heard that no two snowflakes are alike. Your task is to write a program to determine wh ...

  3. 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 ...

  4. 【POJ3349 Snowflake Snow Snowflakes】【Hash表】

    最近在对照省选知识点自己的技能树 今天是Hash 题面 大概是给定有n个6元序列 定义两个序列相等 当两个序列各自从某一个元素开始顺时针或者逆时针旋转排列能得到两个相同的序列 求这n个6元序列中是否有 ...

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

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

  6. POJ 3349 Snowflake Snow Snowflakes (Hash)

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

  7. POJ 3349 Snowflake Snow Snowflakes

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

  8. 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 ...

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

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

随机推荐

  1. 【java JVM】JVM中类的加载,加载class文件的原理机制

    暂时贴图一张,以后补充 解释: 1.符号引用替换为直接引用[参考:http://blog.csdn.net/maerdym/article/details/8087620] 在java中,一个java ...

  2. 使用ab.exe监测100个并发/100次请求情况下同步/异步访问数据库的性能差异

    ab.exe介绍 ab.exe是apache server的一个组件,用于监测并发请求,并显示监测数据 具体使用及下载地址请参考:http://www.cnblogs.com/gossip/p/439 ...

  3. Mac的SourceTree安装时如何注册?

    目前安装SourceTree的过程中是必须注册的,但是注册又是国外的,因此需要FQ.但是未注册成功,是没有办法设置SourceTree的代理的,虽然不知道SourceTree有没有设置代理的功能,解决 ...

  4. fl2440字符设备led驱动

    首先要明白字符设备驱动注册的基本流程 当我们调用insomd命令加载驱动后,驱动程序从module_init函数开始执行:硬件初始化 -> 申请主次设备号 -> 定义fops(file_o ...

  5. HTC相关开发所需SDK等工具都在这里了。 【转】

    OpenVR SDK https://github.com/ValveSoftware/openvr  OpenVR SDK是由原本的SteamWorks SDK更新而来,新增对HTC VIVE开发者 ...

  6. Solr6.6 配置中文分词库mmseg4j

    1.准备 首先安装solr:参照搜索引擎Solr-6.6.0搭建,如果版本高于6,可能会不支持,需要改mmseg4j包 mmseg4j包下载: mmseg4j-solr-2.3.0-with-mmse ...

  7. 为甚么要将某个方法声明为final呢?

    他可以防止其他人覆盖该方法.但更重要的一点或许是:这样做可以有效的"关闭"动态绑定,或者说, 告诉编译器不需要对其进行丰台绑定.这样,编译器就可以为final方法调用生成更有效的代 ...

  8. 使用UnityVS1.8.2搭配VS2013进行Unity3D代码调试(下载+安装+使用+问题解决)

    http://blog.dou.li/UnityVS1_8_2-VS2013.html 备整合UnityVS1.8.2和VS2013调试Unity3D代码,安装后会有些问题无法调试,详情继续往下看. ...

  9. Json_decode:详解

    Json_decode:详解 json_decode - 对 JSON 格式的字符串进行编码   mixed json_decode ( string $json [, bool $assoc = f ...

  10. android Menory 小结

    不建议在Activity中使用static 变量,考虑使用Application.当然,static final例外 但Application也不要cache某个Activity使用的View,如果c ...