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。
#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
- 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 ...
- 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 ...
- 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 ...
- 【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 ...
- POJ 3349:Snowflake Snow Snowflakes(数的Hash)
http://poj.org/problem?id=3349 Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K T ...
- POJ 3349 Snowflake Snow Snowflakes (Hash)
Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 48646 Accep ...
- POJ 3349 Snowflake Snow Snowflakes
Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 27598 Accepted: ...
- 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 ...
- 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
- RxJava 1.x 理解-3
在 RxJava 1.x 理解-1 中,我们说到了RxJava的简单用法,但是这还远远不够,因为 输入的数据 ---> 被监听者(订阅源)对这些数据进行操作,或者执行响应的处理 --> 产 ...
- OpenDigg - 挖掘优质开源项目库
OpenDigg - 挖掘优质开源项目库 OpenDigg专注于挖掘优质的开源项目库,通过技术和人工将软件项目分类整理,同时辅助简要的编译,方便广大程序员便捷地找到需要的开源项目. OpenDigg刚 ...
- CASS 7.1 和 AutoCAD 2006的安装使用
CAD 2006由于是一个古老的版本,所以在WIN7,WIN10上直接安装的话,一般无法成功.此外,AutoCAD 2006是不分32位64位的,之后的版本都是区分的. 但是,对于我这种几年不用一次C ...
- Android studio百度地图demo出现230错误,key校验失败
转自daoxiaomianzi原文 Android studio 百度地图demo出现230错误,key校验失败 使用AndroidStudio导入Baidu地图的as版的demo,引入后,发现没有k ...
- ASP.NET Core 1.0基础之诊断
来源https://docs.asp.net/en/latest/fundamentals/diagnostics.html ASP.NET Core 1.0包含了一些新的特性来辅助诊断问题.可以在S ...
- 搭建vue全家桶
1.直接利用vue-cli脚手架快速搭建 (1)全局安装vue-cli npm install -g vue-cli (2)创建项目 vue init webpack-simple my-projec ...
- SONY 手提 realtek high definition audio driver
SONY 手提 realtek high definition audio driver: 官方网站下载:http://www.realtek.com.tw/downloads/downloadsVi ...
- ajax请求不能下载文件(转载)
最近在做文件下载,后台写了个控制层,直接走进去应该就可以下载文件,各种文件图片,excel等 但是起初老是下载失败,并且弹出下面的乱码: 前台请求代码: $('#fileexcel').unbind( ...
- MongoDB集群设置集合分片生效及查看集合分片情况
#连接mongos /opt/mongodb/mongodb-linux-x86_64-2.4.8/bin/mongo 127.0.0.1:27017 #使用admin数据库 use admin ...
- CMA连续物理内存用户空间映射---(二)
摘要: 相对于上一篇測试程序CMA连续物理内存用户空间映射---(一) 添加功能: 1.分配和映射统一放在IOCTL,一次完毕,能够连续多次分配并映射到用户空间,提高操作性: 2.驱动添加链表,使分配 ...