zoukankan      html  css  js  c++  java
  • POJ 3349-Snowflake Snow Snowflakes

    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
    .
    .
    .
    .
    .

    程序:
    #include<iostream>
    #include<stdio.h>
    #include<string.h>
    using namespace std;
    int n,tot,p=99991,snow[200001][6],head[200001],next[200001];
    
    int hash(int *a)
    {
        int sum=0,mul=1;
        for (int i=0;i<6;i++)
        {
            sum=(sum+a[i])%p;
            mul=(long long)mul*a[i]%p;
        }
        return (sum+mul)%p;
    }
    
    bool check(int *a,int *b)
    {
        for (int i=0;i<6;i++)
            for (int j=0;j<6;j++)
            {
                bool eq=true;
                for (int k=0;k<6;k++)
                    if (a[(i+k)%6]!=b[(j+k)%6]) eq=false;
                if (eq==true) return true;
                eq=true;
                for (int k=0;k<6;k++)
                    if (a[(i+k)%6]!=b[(j-k+6)%6]) eq=false;
                if (eq==true) return true;
            }
        return 0;
    }
    
    bool insert(int *a)
    {
        int val=hash(a);
        for (int i=head[val];i;i=next[i])
            if (check(snow[i],a)) return 1;
        tot++;
        memcpy(snow[tot],a,6*sizeof(int));
        next[tot]=head[val];
        head[val]=tot;
        return 0;
    }
    
    int main()
    {
        cin>>n;
        for (int i=1;i<=n;i++)
        {
            int a[10];
            for (int j=0;j<6;j++) scanf("%d",&a[j]);
            if (insert(a))
            {
                puts("Twin snowflakes found.");
                return 0;
            }
        }
        puts("No two snowflakes are alike.");
        return 0;
    }
    
  • 相关阅读:
    【二】调通单机版的thrift-C++版本
    【一】调通单机版的thrift-python版本
    Spark在实际项目中分配更多资源
    Spark实际项目中调节并行度
    IDEA中大小写转换快捷键
    使用maven下载cdh版本的大数据jar包
    【Hive六】Hive调优小结
    【Hive五】Hive函数UDF
    【Hbase三】Java,python操作Hbase
    【Hive三】Hive理论
  • 原文地址:https://www.cnblogs.com/YYC-0304/p/9499893.html
Copyright © 2011-2022 走看看