Poj 3349 Snowflake Snow Snowflakes(哈希+vector)

Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 34759		Accepted: 9123

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.

刚上来我写了，一个二分但WA。。。我感觉是没有考虑顺序问题，只是判断了数值上满足就够了

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#define LL long long
#define mod 100001
using namespace std;
struct node
{
    int a[8];
};
vector<node>ha[200001];
bool ju(node a,node b)
{
    int i;
    for(i=0;i<6;i++)//分别判断顺时针和逆时针是否相同<span id="transmark"></span>
    {
        if(a.a[0]==b.a[i]
        &&a.a[1]==b.a[ (i+1)%6]
        &&a.a[2]==b.a[(i+2)%6]
        &&a.a[3]==b.a[(i+3)%6]
        &&a.a[4]==b.a[(i+4)%6]
        &&a.a[5]==b.a[(i+5)%6])
        return true;
        if(a.a[0]==b.a[i]
        &&a.a[1]==b.a[(i+5)%6]
        &&a.a[2]==b.a[(i+4)%6]
        &&a.a[3]==b.a[(i+3)%6]
        &&a.a[4]==b.a[(i+2)%6]
        &&a.a[5]==b.a[(i+1)%6])
        return true;
    }
    return false;
}
int main()
{
    node q;
    LL n,s,ans;
    scanf("%lld",&n);
    for(int i=0;i<100001;i++)//使用vector前要清空
        ha[i].clear();
    ans=0;//标记是否找到符合条件的序列
    for(int i=0;i<n;i++)
    {
        s=0;
        for(int j=0;j<6;j++)
        {
            scanf("%d",&q.a[j]);
            s+=q.a[j];//将当前所有数的和当作哈希头
        }
        if(ans==-1)
        continue;
        s%=mod;
        if(ha[s].size()==0)//若为空直接进
            ha[s].push_back(q);
        else
        {
            for(int j=0;j<ha[s].size();j++)//否则的话，比较ha[s]已有的和当前的序列是否可能相同
            {
                if(ju(q,ha[s][j])==true)
                {
                    ans=-1;
                    break;
                }
            }
            if(ans==0)//如果没有找到便直接输入
            ha[s].push_back(q);<span id="transmark"></span>
        }
    }
    if(ans==-1)
    puts("Twin snowflakes found.");
    else
    puts("No two snowflakes are alike.");
    return 0;
}