poj3349找相同的雪花（哈希）

题目传送门

题目大意：给你n个雪花，每个雪花的六个棱都有各自的长度，如果存在两片雪花的每条棱长度对应相同，则输出一句英文，如果不存在就输出另外一句英文，n和长度都比较大。

思路：第一次真正接触哈希，查了题解，还没能通过这道题get到哈希的精髓。这道题是把雪花的六个长度加起来取模的值作为哈希值，然后把拥有相同哈希值的雪花归为一类，每一次输入新的雪花，就先找到它的那一类，然后这一类中进行遍历，遍历的方式也就是在一个环里面暴力，我对这道题用哈希的理解是，利用哈希值进行分类（和并查集有点像），在类中寻找，减少遍历的次数。

代码有详细注释。

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const double PI=acos(-1.0);
int fact[10]= {1,1,2,6,24,120,720,5040,40320,362880};
const int mod = 999983;
const int maxn = 100005;
int tot = 0,n,head[mod+5],v[maxn][6],next[maxn];

int gethash(int id) { //得到哈希值  每位都加起来取余
	int hashval=0;
	for(int i=0; i<6; i++) {
		hashval=(hashval%mod+v[id][i]%mod)%mod;
	}
	return hashval;
}
void addvhead(int hashval) { //邻接表
	next[tot]=head[hashval];//head表示  六位数总和为 hashval的最后一个元素的标号    next[i]表示i的上一个表示是什么
	head[hashval]=tot++;//更新head   tot就是当前加入的id
}
bool compare(int id,int x) {
	for(int i=0; i<6; i++) { //顺时针 比较
		int f=0;
		for(int st=i,j=0; j<6; j++,st= st+1 >= 6 ? 0 : st+1) {
			if(v[id][st]==v[x][j])f++;
			else break;
		}
		if(f==6)return true;
	}
	for(int i=0; i<6; i++) { //逆时针
		int f=0;
		for(int st=i,j=5; j>=0; j--,st= st+1 >= 6 ? 0 : st+1) {
			if(v[id][st]==v[x][j])f++;
			else break;
		}
		if(f==6)return true;
	}
	return false;
}
bool pd(int id) {
	int hashval=gethash(id);
	for(int i=head[hashval]; i!=-1; i=next[i]) { //遍历哈希值相同的雪花
		if(compare(id,i))return true;
	}
	addvhead(hashval);

	return false;
}
int main() {
	memset(head,-1,sizeof(head));
	cin>>n;
	bool flag=false;
	for(int i=0; i<n; i++) {
		for(int j=0; j<6; j++)scanf("%d",&v[i][j]);
		if(flag)continue;
		if(pd(i))flag=true;

	}
	if(flag)cout<<"Twin snowflakes found."<<endl;
	else cout<<"No two snowflakes are alike."<<endl;
}

Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 47318		Accepted: 12345

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 nlines, 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.