哈希—— POJ 3349 Snowflake Snow Snowflakes

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Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 33595		Accepted: 8811

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.

题意：

一片雪花有6片叶子，给出n片雪花，下面每行6个数字。分别代表每片叶子长度。问是否存在有2片雪花形态是同样的。

形态同样的定义：例如以下图，2 3 4 5 6 1 和 4 5 6 1 2 3 和 3 2 1 6 5 4 都是形态同样的雪花，即是同一片雪花。

即是一片雪花能够从某一个数開始顺时针或逆时针数。

思路：

把一片雪花的全部长度相加 mod 一个大质数（100w左右）作为键值key，那就仅仅有key同样的雪花才有可能是形态同样的。

而key同样的雪花会映射到哈希表的同一个槽，那我们用链表把key值同样的在同一个槽连起来。仅仅有key相等的时候就一个个比較key值那条链的雪花是否存在形态同样的，假设有就标记，以后就仅仅是读入数据不比較；假设没有就把雪花增加链表。

也能够採用开方性寻址的方法。思路是一样的。比較雪花形态是否相等要分顺时针和逆时针

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <map>
#include <iostream>
using namespace std;
#define N 100007
const int prime = 999983;

typedef struct
{
	int key;
	int arm[7];
}Point;

Point *slot[prime+10];

bool Check(Point *p, Point *q) //p是新的，q是旧的
{
	int i, j;
	//顺时针
	for(j = 0; j < 6; j++){
		for(i = 0; i < 6; i++)
			if(q->arm[i] != p->arm[(i+j)%6]) break;
		if(6 == i) return true;
	}
	//逆时针
	for(j = 0; j < 6; j++){
		for(i = 0; i < 6; i++){
			if(q->arm[i] != p->arm[(12-i-j)%6]) break;
		}
		if(6 == i) return true;
	}
	return false;
}

int Hash(int k)
{
	/*
	char s[20];
	sprintf(s, "%d", k);
	int i, h = 0, a = 232;
	for(i = 0; i < strlen(s); i++)
		h = (a * h + s[i] - '0') % prime;
		*/
	return (k<<2);
}

bool Try_to_insert(Point *p)
{
	int k = p->key;
	if(NULL == slot[k]){ //有空槽
		slot[k] = p;
		return true;
	}
	while(slot[k])
	{
		if(Check(p, slot[k])) return false;
		k = (k + Hash(k)) % prime;
	}
	slot[k] = p;
	return true;
}

int main()
{
	//freopen("in.txt","r",stdin);
	int i, j, n, flag = 0;
	Point *p;
	scanf("%d", &n);
	for(i = 0; i < n; i++){
		p = (Point *)malloc(sizeof(Point));
		p->key = 0;
		for(j = 0; j < 6; j++){
			scanf("%d", &p->arm[j]);
			p->key = (p->key + p->arm[j]) % prime;
		}
		if(flag) continue;
		if(!Try_to_insert(p)) flag = 1;
	}
	if(flag) printf("Twin snowflakes found.\n");
	else printf("No two snowflakes are alike.\n");
	return 0;
}