POJ3349-Snowflake Snow Snowflakes(hash表)

POJ3349 原题链接:http://poj.org/problem?id=3349

Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 43393		Accepted: 11385

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.

题目大意:输入n片雪花,一片雪花有6片花瓣,每片花瓣的形状用数字表示,判断是否存在相同的雪花;这六个整数是从任意一个分支开始,朝顺时针或逆时针方向遍历得到的。

思路:通过这道题学习到了hash表,这道题最简单的思路就是枚举任两个雪花进行比较看是否存在相同的雪花,但是这样时间复杂度为O(n^2),那么有没有时间较短的方法呢,那就是哈希表,哈希表的基本原理我放在我的另一篇博客里,这里简单的说一下:

1.定义一个hash数组hash[key][i]与一个雪花数组snow[n][7]; hash[key][i]的值表示为key类雪花中的第i个雪花在snow[n]中的下标

2.将雪花按某种方式分类,我这里是定义一个雪花的key值为6片花瓣的和*2/prime=99991 即:(∑(snow[i][j]) j=1~6)*2/prime;

3.在读入一个雪花的时候把这些情况全部放入哈希表中，如果某次插入的时候发生冲突，则说明存在重复的雪花.

代码如下:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector>
using namespace std;
const int max_size=100010;
const int prime=99991;
int **hsh=new int*[max_size];  //二维动态数组节省空间
int key_len[max_size];
int snow[max_size][7];
int Compare(int a,int b)
{
    for(int i=0;i<6;i++){
        if((
            snow[a][0]==snow[b][i]&& snow[a][1]==snow[b][(i+1)%6]&&
            snow[a][2]==snow[b][(i+2)%6]&& snow[a][3]==snow[b][(i+3)%6]&&
            snow[a][4]==snow[b][(i+4)%6]&& snow[a][5]==snow[b][(i+5)%6])
        ||                                                                          //判断两片雪花是否相同
        (   snow[a][0]==snow[b][i]&& snow[a][1]==snow[b][(i+5)%6]&&
            snow[a][2]==snow[b][(i+4)%6]&& snow[a][3]==snow[b][(i+3)%6]&&
            snow[a][4]==snow[b][(i+2)%6]&& snow[a][5]==snow[b][(i+1)%6]))
        return 1;
    }
    return 0;
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int sum=0;
        int key;
        for(int j=0;j<6;j++)
        {
            scanf("%d",&snow[i][j]);
            sum+=snow[i][j]*2;
        }
        key=sum%prime;   //将雪花进行分类
        for(int k=1;k<=key_len[key];k++)
        {
            if(Compare(i,hsh[key][k])==1)
            {
                cout<<"Twin snowflakes found.\n"<<endl;
                exit(0);
            }
        }
        key_len[key]++;
        hsh[key]=new int[key_len[key]];   //将key类雪花的下标i存储进hash数组
        hsh[key][key_len[key]]=i;
    }
    printf("No two snowflakes are alike.\n");
}