POJ 3349:Snowflake Snow Snowflakes(数的Hash)
http://poj.org/problem?id=3349
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 37609 | Accepted: 9878 |
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.
#include<cstdio>
#include<algorithm>
#include <cstring>
#include <iostream>
using namespace std;
#define MOD 14997
struct node
{
int len[];
}snow,f[][];
int Hash[]; /*
第一道Hash
判断两个雪花是否相同:
分别是顺时针旋转和逆时针旋转,只要有一次是满足条件的就可以当成是一样的
一开始MOD取了999983,结果MLE了,后来看了一下别人的才取了14997
*/
bool check(node a,node b)
{
int i,j,k;
bool flag;
//顺时针旋转判断
for(i=;i<;i++){
flag=true;
for(j=i,k=;k<;k++,j=(j+)%){
if(a.len[j]!=b.len[k]){
flag=false;break;
}
}
if(flag) return ;
}
//逆时针旋转判断
for(i=;i<;i++){
flag=true;
for(j=i,k=;k<;k++,j=(j+)%){
if(a.len[j]!=b.len[k]){
flag=false;break;
}
}
if(flag) return ;
}
return ;
} int main()
{
int n;
int now;
scanf("%d",&n);
//只有一组样例就不初始化了
bool flag=false;
for(int i=;i<n;i++){
now=;
for(int j=;j<;j++){
scanf("%d",&snow.len[j]);
now+=snow.len[j]%MOD;
}
if(flag) continue;
now%=MOD;
if(Hash[now]>){
for(int j=;j<=Hash[now];j++){
if(check(snow,f[now][j])){
flag=true;break;
}
}
}
Hash[now]++;
f[now][Hash[now]]=snow;
}
if(flag) printf("Twin snowflakes found.\n");
else printf("No two snowflakes are alike.\n");
return ;
}
2016-06-04
POJ 3349:Snowflake Snow Snowflakes(数的Hash)的更多相关文章
- POJ 3349 Snowflake Snow Snowflakes (Hash)
Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 48646 Accep ...
- 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 ...
- POJ 3349 Snowflake Snow Snowflakes
Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 27598 Accepted: ...
- [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 ...
- 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 ...
- 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 ...
- [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. ...
- POJ 3349 Snowflake Snow Snowflakes Hash
题目链接: http://poj.org/problem?id=3349 #include <stdio.h> #include <string.h> #include < ...
- 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 ...
随机推荐
- MVC 应用程序级别捕捉异常
捕捉异常: using System; using System.IO; using System.Net; using System.Net.Http; using System.Net.Http. ...
- [3D]绘制线
数据实体: using System; using System.Collections.Generic; using System.Linq; using System.Text; using Sy ...
- PLSQL 申明和游标
--从键盘输入一个数 accept b prompt '请输入一个大于零的数字'; declare anum number := &b; begin loop dbms_output.put_ ...
- Oracle 监听器日志文件过大导致监听异常
Oracle 监听器日志文件过大导致监听异常 db版本:11.2.0.1 os版本:windows2008 现象: 应用异常,无法连接数据库.登陆数据库服务器,查看监听已经断掉.尝试重启监听,重启失败 ...
- jQuery 扩展功能
源码如下: /*! * 说明:Jquery库扩展 * 创建时间: leo 2016/10/13 */ (function (window, jQuery, undefined) { jQuery.ex ...
- J2EE sitemesh使用
maven包含sitemesh: <dependency> <groupId>opensymphony</groupId> <artifactId>si ...
- Compile Sources 和 Copy Bundle Resources的区别
Compile Sources主要存放.m文件 Copy Bundle Resources 主要存放xib plist bundle js 文件
- nyist 676 小明的求助
http://acm.nyist.net/JudgeOnline/problem.php?pid=676 小明的求助 时间限制:2000 ms | 内存限制:65535 KB 难度:2 描述 ...
- poj 题目分类(2)
初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj329 ...
- AR 应收 表
AR 应收 应收事务处理相关表 SELECT * FROM ar.ar_batches_all; --事务处理批 SELECT * FROM ar.ra_custom ...