poj -- 3349--Snowflake Snow Snowflakes (暴力)

本文讨论了如何通过输入多个雪花的臂长数据,判断是否存在两个完全相同的雪花,包括直接比较和考虑同构的情况。


Snowflake Snow Snowflakes
Time Limit: 4000MS Memory Limit: 65536K
Total Submissions: 18980 Accepted: 4926

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:题目的意思是给出N个雪花片,每个雪花片有6个边,问存不存在2个完全相同的雪花片


此题测试数据有问题,因此很多人直接排序就过了


这道题还要考虑是不是同构的问题

2

1 2 3 4 5 6

3 1 2 4 6 5

照道理说应该输出:No two snowflakes are alike.

因此排序之后还要比较雪花片原来的样子是不是符合同构。

#include<iostream>
#include <ctime>
#include <algorithm>
using namespace std;

int num[100010][12];
bool judge;
int cmp(const void *c,const void *d)
{
	int *p=(int*)c;
	int *q=(int*)d;
	sort(p,p+6);
	sort(q,q+6);
	int i;
	for( i=0;i<6;i++)
		if(p[i]!=q[i])
			return p[i]-q[i];
	if(i==6)
	{
		return 0;
	}
}
bool cmp2(int *p,int *q)
{
	if(p[0]==q[0]&&p[1]==q[1]&&p[2]==q[2]&&p[3]==q[3]&&p[4]==q[4]&&p[5]==q[5])
		return true;
	if(p[0]==q[1]&&p[1]==q[2]&&p[2]==q[3]&&p[3]==q[4]&&p[4]==q[5]&&p[5]==q[0])
		return true;
	if(p[0]==q[2]&&p[1]==q[3]&&p[2]==q[4]&&p[3]==q[5]&&p[4]==q[0]&&p[5]==q[1])
		return true;
	if(p[0]==q[3]&&p[1]==q[4]&&p[2]==q[5]&&p[3]==q[0]&&p[4]==q[1]&&p[5]==q[2])
		return true;
	if(p[0]==q[4]&&p[1]==q[5]&&p[2]==q[0]&&p[3]==q[1]&&p[4]==q[2]&&p[5]==q[3])
		return true;
	if(p[0]==q[5]&&p[1]==q[0]&&p[2]==q[1]&&p[3]==q[2]&&p[4]==q[3]&&p[5]==q[4])
		return true;
	if(p[0]==q[0]&&p[5]==q[1]&&p[4]==q[2]&&p[3]==q[3]&&p[2]==q[4]&&p[1]==q[5])
		return true;
	if(p[0]==q[1]&&p[5]==q[2]&&p[4]==q[3]&&p[3]==q[4]&&p[2]==q[5]&&p[1]==q[0])
		return true;
	if(p[0]==q[2]&&p[5]==q[3]&&p[4]==q[4]&&p[3]==q[5]&&p[2]==q[0]&&p[1]==q[1])
		return true;
	if(p[0]==q[3]&&p[5]==q[4]&&p[4]==q[5]&&p[3]==q[0]&&p[2]==q[1]&&p[1]==q[2])
		return true;
	if(p[0]==q[4]&&p[5]==q[5]&&p[4]==q[0]&&p[3]==q[1]&&p[2]==q[2]&&p[1]==q[3])
		return true;
	if(p[0]==q[5]&&p[5]==q[0]&&p[4]==q[1]&&p[3]==q[2]&&p[2]==q[3]&&p[1]==q[4])
		return true;
	return false;
}
int main()
{
	int k=0,i,j,t,n;
//	freopen("a.out","r",stdin);
	double q=clock();
	while(scanf("%d",&n)!=EOF)
	{
		judge=false;
		for(j=0;j<n;j++)
			for(i=0;i<6;i++)
			{
				scanf("%d",&num[j][i]);
				num[j][i+6]=num[j][i];
			}
		qsort(num,n,sizeof(num[0]),cmp);
		for(j=1;j<n;j++)
		{
			if(cmp2(num[j]+6,num[j-1]+6))
				judge=true;
		}
		if(judge)
			printf("Twin snowflakes found.\n");
		else
			printf("No two snowflakes are alike.\n");
	}
	//printf("%lf\n",clock()-q);
	
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值