POJ 3349, Snowflake Snow Snowflakes

hash table


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.

Source
CCC 2007


// POJ3349.cpp : Defines the entry point for the console application.
//

#include 
<iostream>
using namespace std;

struct Snowflake
{
    Snowflake():next(
0){}
    Snowflake
* next;
    
int arms[6];
};
bool compare (const Snowflake& sf1, const Snowflake& sf2)
{
    
// left to right
    for (int j = 0; j < 6++j)    
        
if (sf1.arms[0== sf2.arms[j])
        {
            
int k = 0;
            
for (; k < 6++k)
            {
                
if (sf1.arms[k] != sf2.arms[(j + k) % 6]) break;
            };
            
if (k == 6return true;
        };
    
//right to left
    for (int j = 0; j < 6++j)    
        
if (sf1.arms[0== sf2.arms[j])
        {
            
int k = 0;
            
for (; k < 6++k)
            {
                
if (sf1.arms[k] != sf2.arms[(j - k + 6% 6]) break;
            };
            
if (k == 6return true;
        };
    
return false;
}

int main(int argc, char* argv[])
{
    
int N;
    scanf(
"%d\n"&N);

    
const int SIZE = 33119;
    Snowflake hash[SIZE];
    memset(hash, 
0sizeof(hash));
    
bool found = false;
    
for (int i = 0; i < N; ++i)
    {
        Snowflake
* sf = new Snowflake;
        
for (int j = 0; j < 6++j)scanf("%d"&sf->arms[j]);
        
if (found == false)
        {
            
int key = sf->arms[0+ sf->arms[1+ sf->arms[2+ sf->arms[3+ sf->arms[4+ sf->arms[5];

            Snowflake
* pt = &hash[key%SIZE];
            
while( pt ->next != NULL){
                pt 
= pt ->next;
                
if (compare(*pt,*sf)) found = true;
            }
            pt
->next = sf;
        }
    };
    
if (found) cout << "Twin snowflakes found.\n";
    
else cout << "No two snowflakes are alike.\n";
    
return 0;
}

原文地址:https://www.cnblogs.com/asuran/p/1578949.html