poj 3349 Snowflake Snow Snowflakes

最新推荐文章于 2020-09-06 17:13:10 发布

原创最新推荐文章于 2020-09-06 17:13:10 发布 · 212 阅读

0 ·

CC 4.0 BY-SA版权

hash 专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一个程序设计问题，即如何通过比较六边形雪花的边长来判断它们是否完全相同。通过对输入数据进行哈希处理并采用两种方向的比较方法，确保能够准确地识别出可能相同的雪花。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 38481		Accepted: 10101

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

题目大意：给定雪花的六条边长，不论顺序，只要其中有两个雪花六条边的长度都相等，那么输出“Twin snowflakes found”，否则输出“No two snowflakes are alike.”

#include<iostream>
#include<cstring>
#include<cstdio>
#define N 30010
using namespace std;

struct node
{
    int record[6];
}temp, mould[15000][20];
int n, m;
int pos;
int num1[15000];

int hash( node x )            //进行hash
{
    int sum = 0;
    for ( int i = 0;i < 6; i++ )
    {
        sum = sum + x.record[i];
    }
    return sum%14997;
}

int compare( node x, node y )
{
    for ( int i = 0;i < 6; i++ )                     //顺时针比较
    {
        int j = 0;
        int k = i;
        while ( x.record[k] == y.record[j])
        {
            k++;
            k = k%6;
            j++;
        }
        if( j == 6 )
            return 1;
    }
    for ( int i = 0;i < 6; i++ )             //逆时针比较
    {
        int j = 0;
        int k = i;
        while ( x.record[k] == y.record[j])
        {
            k++;
            k = (k+4)%6;
            j++;
        }
        if( j == 6 )
            return 1;
    }
    return 0;
}

int main()
{
    int j;
    scanf( "%d", &n );
    for ( int i = 0;i < n; i++ )
    {
        for ( j = 0;j < 6; j++)
        {
            scanf ( "%d", &temp.record[j] );  //记录长度
        }
        pos = hash( temp );                        //进行hash
        for ( j = 0;j < num1[pos]; j++ )
        {
            if ( compare( temp, mould[pos][j] ) )            //如果有相同的
            {
                printf("Twin snowflakes found.\n");
                 return 0;
            }
        }
        mould[pos][j] = temp;                          //如果不相同就++；
        num1[pos]++;
    }
    printf("No two snowflakes are alike.\n");
    return 0;
}

代码菜鸟，如有错误，请多包涵！！！