POJ 3349 Snowflake Snow Snowflakes

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 byn 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

21 2 3 4 5 64 3 2 1 6 5

Sample Output

Twin snowflakes found.

Source

CCC 2007

/*
   给定N个雪花的边长，问是否有两个相同的雪花，直接用链表哈希，不过链表千万不要开的太大会导致遍历的时候超时。

   方法：两个雪花相同的前提是他们6个边的边长之和肯定是相同的。而又由于两个雪花给定边长的顺序不同，所以没两片雪花要比对6*2遍。

*/

#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstring>#include <string>#include <algorithm>const int INF = 0x3f3f3f3f;using namespace std;struct node{    int sh[6];    node *next;}*head[10010];//10007bool f(int *u,int *v) //比对边长是否相同{    int i,j,k;    for(int i=0; i<6; i++)    {        for( j=(i-1+6)%6, k=1 ; k<6 && u[k]==v[j]; j=((--j)+6)%6, k++);              if( k==6 ) return true;        for( j=(i+1)%6 ,k=1; k<6 && u[k]==v[j]; j=(++j)%6 ,k++) ;              if( k==6) return true;    }    return false;}int main(){    int n;    node *p,*q;    for(int i=0;i<=10009;i++)    {        head[i] = new node;        head[i] -> next = NULL;    }    while(~scanf("%d",&n))    {        for(int i=0; i<n; i++)        {            p = new node;            p->next = NULL;            int sum = 0;            for(int j=0; j<6; j++)            {                scanf("%d",&p->sh[j]);                sum+=p->sh[j];            }            if(head[sum%10007]->next == NULL)            {                head[sum%10007]->next = p;            }            else            {                p->next = head[sum%10007]->next;                head[sum%10007]->next = p;            }        }        bool Flag = false;        for(int i=0; i<=10007; i++)        {            if(head[i]->next!=NULL && head[i]->next->next!=NULL)            {                p = head[i]->next;                while(p)                {                    q = p->next;                    while(q)                    {                        if(f(p->sh,q->sh))                        {                            Flag = true;                            break;                        }                        q=q->next;                    }                    if(Flag)                        break;                    p=p->next;                }            }            if(Flag)                break;            head[i]->next = NULL;        }        if(Flag)            printf("Twin snowflakes found.\n");        else            printf("No two snowflakes are alike.\n");    }    return 0;}