[POJ 3349]Snowflake Snow Snowflakes

Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 28891		Accepted: 7622

 

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：

给你n个雪花，每个雪花有六个花瓣代表六个值，判断是否有至少一对相同的雪花。（可以顺时针也可以逆时针，可以从任意一个角开始，具体看样例）

题解：

由于n较大，一个一个去比较的暴力算法肯定会超时，那么我们考虑hash查找。

由于n<=100000，所以最多的情况就是有100000个不同的雪花，那么我们可以将六个角的长度之和对100000取模得到键值，然后如果键值相同就用一个链表储存起来，然后查找的时候只需要查找键值相同的就可以了。

由于题目所问的是有无相同的雪花，所以只要找到了，后面的就没有必要再找了。

代码如下：

 

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const int mod=100005;
int n;
bool ok;
struct node
{
    int a[7];
}a[100005];
struct hash
{
    int next,key,place;
}edge[100005];
int head[100005],size;
bool compare(node x,node y)
{  
    sort(x.a+1,x.a+7);
    sort(y.a+1,y.a+7);
    for(int i=1;i<=6;i++)  
    {
        if(x.a[i]!=y.a[i])
            return false;
    }
    return true;
}
void push_hash(int k,int p)
{
    int key=(k%mod+mod)%mod;
    size++;
    edge[size].key=k;
    edge[size].place=p;
    edge[size].next=head[key];
    head[key]=size;
}
void find_hash(int k,node v)
{
    int key=(k%mod+mod)%mod;
    for(int i=head[key];i!=-1;i=edge[i].next)
    {
        if(compare(a[edge[i].place],v))
        {
            ok=1;
            return;
        }
    }
}
int main()
{
    memset(head,-1,sizeof(head));
    int i,j,t;
    scanf("%d",&n);
    for(t=1;t<=n;t++)
    {
        int key=0;
        for(i=1;i<=6;i++)
        {
            scanf("%d",&a[t].a[i]);
            key+=a[t].a[i];
        }
        if(!ok)
        {
            find_hash(key,a[t]);
            push_hash(key,t);
        }
    }
    if(ok)printf("Twin snowflakes found.\n");
    else printf("No two snowflakes are alike.\n");
    return 0;
}