POJ3349 Snowflake Snow Snowflakes (hash
Snowflake Snow Snowflakes
Time Limit: 4000MS | Memory Limit: 65536K | |
Total Submissions: 48624 | Accepted: 12697 |
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. 代码如下:
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e5+;
int n,tot,P=;
int snow[maxn][];
int head[maxn];
int nex[maxn];
int H(int *a){
int sum=;
int mul=;
for(int i=;i<;i++){
sum=(sum+a[i])%P;
mul=(long long)mul*a[i]%P;
}
return (sum+mul)%P;
}
bool equal(int *a,int *b){
for(int i=;i<;i++){
for(int j=;j<;j++){
bool eq=;
for(int k=;k<;k++){
if(a[(i+k)%]!=b[(j+k)%]) eq=;
}
if(eq) return ;
eq=;
for(int k=;k<;k++){
if(a[(i+k)%]!=b[(j-k+)%]) eq=;
}
if(eq) return ;
}
}
return ;
} bool insert(int *a){
int val=H(a);
for(int i=head[val];i;i=nex[i]){
if(equal(snow[i],a)) return ;
}
++tot;
memcpy(snow[tot],a,*sizeof(int));
nex[tot]=head[val];
head[val]=tot;
return ;
}
int main(){
cin>>n;
for(int i=;i<=n;i++){
int a[];
for(int j=;j<;j++) scanf("%d",&a[j]);
if(insert(a)){
puts("Twin snowflakes found.");
return ;
}
}
puts("No two snowflakes are alike.");
}
最新文章
- 为jQuery的$.ajax设置超时时间
- laravel 的.env 配置文件
- [MSSQL2012]LEAD函数
- Struts2框架的运行流程
- javascript图片预先加载
- 如何实现带照片缩略图的Listview
- Yii框架中的CURD操作
- NSURL 子串截取
- MySQL ibdata多路径扩容
- vue 父组件向子组件传递事件/调用事件
- Flink中的Time
- CentOS 7的安装详解
- McQueenRPC源码阅读
- 20165306 2017-2018-2《Java程序设计》课程总结
- hasura graphql-engine v1.0.0-alpha30 版本新功能介绍
- 9.11 h5日记
- Java数组的定义和使用
- Git之代码合并及分支管理
- struts2訪问servlet的API
- C++设计模式之享元模式