poj 3349(hash判重)

最新推荐文章于 2020-03-07 11:34:01 发布

转载最新推荐文章于 2020-03-07 11:34:01 发布 · 66 阅读

本文介绍了一种用于比对雪flake是否相同的算法实现。通过旋转和翻转雪flake的六边形结构，寻找字典序最小的状态进行哈希存储，以此来判断是否有完全相同的两个雪flake。该算法在比赛中获得了成功应用。

挺有意思的题。

思路不难，用hash判重，但是要坑人的是一开始怎么将每串序列按照同一规律排列。

比如

2 1 2 1 1 2

1 1 2 2 1 2

怎么转化为 1 1 2 2 1 2 ，也就是转化为字典序最小, 最后没有办法只有枚举， 3000+ms 险过.

Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 26254		Accepted: 6900

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.

Source

CCC 2007

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define N 1010000
#define MOD 1000007
#define INF 10000001
typedef __int64 LL;

struct node
{
    int next;
    int g[6];
}edge[N];

int cnt,pre[N];
int flag1;


void hash(LL key,int k[6])
{
    for(int p=pre[key];p!=-1;p=edge[p].next)
    {
        int ff=0;
        for(int i=0;i<6;i++)
            if(k[i]!=edge[p].g[i])
            {
                ff=1;
                break;
            }
        if(ff==0)
        {
            flag1=1;
            return ;
        }
    }
    for(int i=0;i<6;i++)
        edge[cnt].g[i]=k[i];
    edge[cnt].next=pre[key];
    pre[key]=cnt++;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int k[6],tk[6],mk[6];
        cnt=0;
        memset(pre,-1,sizeof(pre));
        flag1=0;
        for(int i=0;i<n;i++)
        {
            for(int i=0;i<6;i++)
            mk[i]=INF;
            for(int j=0;j<6;j++)
            {
                scanf("%d",&k[j]);
            }
            if(flag1==1) continue;
            /*int flag=1;
            int t1=s,t2=s;
            for(int j=0;j<5;j++)
            {
                t1++;
                if(t1>=6) t1-=6;
                t2--;
                if(t2<0) t2+=6;
                if(k[t1]>k[t2])
                {
                    flag=-1; // 说明是逆向的
                    break;
                }
                if(k[t1]<k[t2])
                {
                    flag=1;
                    break;
                }
            }
            for(int j=0;j<6;j++)
            {
                tk[j]=k[s];
                s += flag;
                if(s>=6) s-=6;
                if(s<0) s+=6;
            }*/
            for(int j=0;j<6;j++)
            {
                for(int i1=0;i1<6;i1++)
                    tk[i1]=k[(i1+j)%6];
                int f=0;
                for(int i1=0;i1<6;i1++)
                {
                    if(mk[i1]<tk[i1])
                    {
                        break;
                    }
                    if(mk[i1]>tk[i1])
                    {
                        f=1;
                        break;
                    }
                }
                if(f==1)
                {
                    for(int i1=0;i1<6;i1++)
                        mk[i1]=tk[i1];
                }
            }
            for(int j=0;j<6;j++)
            {
                for(int i1=0;i1<6;i1++)
                    tk[i1]=k[(j-i1+6)%6];
                int f=0;
                for(int i1=0;i1<6;i1++)
                {
                    if(mk[i1]<tk[i1])
                    {
                        break;
                    }
                    if(mk[i1]>tk[i1])
                    {
                        f=1;
                        break;
                    }
                }
                if(f==1)
                {
                    for(int i1=0;i1<6;i1++)
                        mk[i1]=tk[i1];
                }
            }
            LL tmp=0;
            for(int j=0;j<6;j++)
            {
                tmp=(tmp*100000007+mk[j])%MOD;
            }
            hash(tmp,mk);
        }
        if(flag1==0) printf("No two snowflakes are alike.\n");
        else printf("Twin snowflakes found.\n");
    }
    return 0;
}