POJ 3349 Snowflake Snow Snowflakes(排序)

Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 34160		Accepted: 8959

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.

刚学做过的一道题，不记得以前怎么做的了（好像比这个简单、、），问两片雪花是否相同，给定6个整数来描述一个雪花

如a,b,c,d,e,f , 一个雪花可以有12种描述（从每一个数可以顺时针、逆时针一圈）

如以c为起点的两种相同描述为 c,d,e,f,a,b  和 c,b,a,f,e,d

可以先找到每种雪花的12中描述，进行排序取字典序最小的存起来，再将所有雪花的最小字典序进行排序，看相邻是否有相同的

#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
#include <cmath>
#define N 100010
using namespace std;
struct node
{
    int s[6];
} q[N];
int cmp(const void *a,const void *b)
{
    for(int i=0; i<6; i++)
        if((*(node *)a).s[i] != (*(node *)b).s[i] )
            return (*(node *)a).s[i] - (*(node *)b).s[i] ;
    return 0;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int w=0; w<n; w++)
        {
            node a;
            scanf("%d%d%d%d%d%d",&a.s[0],&a.s[1],&a.s[2],&a.s[3],&a.s[4],&a.s[5]);
            node v[12];
            for(int i=0; i<12; i++)        //取出12种描述
            {
                int  t=0;
                if(i<6)
                {
                    for(int j=i; j<6; j++)
                        v[i].s[t++]=a.s[j];
                    for(int j=0; j<i; j++)
                        v[i].s[t++]=a.s[j];
                }
                else
                {
                    for(int j=i-6; j>=0; j--)
                        v[i].s[t++]=a.s[j];
                    for(int j=5; j>i-6; j--)
                        v[i].s[t++]=a.s[j];
                }

            }
            qsort(v,12,sizeof(v[0]),cmp);    //12种进行排序，取字典序最小的
            q[w]=v[0];
        }
        bool flag = false;
        qsort(q,n,sizeof(q[0]),cmp);
        for(int i=1; i<n; i++)
        {
            bool flag1=true;
            for(int j=0; j<6; j++)
            {
                if(q[i].s[j] != q[i-1].s[j])
                {
                    flag1 = false;
                    break;
                }
            }
            if(flag1)
            {
                flag=true;
                break;
            }
        }
        if(flag) printf("Twin snowflakes found.\n");
        else printf("No two snowflakes are alike.\n");
    }
    return 0;
}