poj-3349 SnowflakeSnowSnowflakes

最新推荐文章于 2021-02-13 22:50:55 发布

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最新推荐文章于 2021-02-13 22:50:55 发布

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文章标签： each integer pair input 64bit search

本文链接：https://blog.youkuaiyun.com/cfm_2914/article/details/7405203

本文介绍了一种用于识别雪花是否相同的算法。通过将雪花的六臂长度数据进行特定处理，并利用哈希表来快速查找可能相同的雪花，实现了高效比对。文章提供了完整的C++实现代码。

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A - SnowflakeSnowSnowflakes

Crawling in process... Crawling failed Time Limit:4000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3349

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.

这是对hash表的应用，先定义一个较大的数，求出每组数据的总和，若和取模这个较大数，若结果相同就放在同一个hash表中，在比较这些组中每个数字是否相同。

#include<vector>
using namespace std;
int d[100001][6];
const int MAXVALUE=99991;
vector<int> hash[MAXVALUE];
bool isSame(int a,int b)
{
    int i; 
    for(i=0;i<6;i++)
    {
        if( ( d[a][0]==d[b][i] && 
        d[a][1]==d[b][(i+1)%6] && 
        d[a][2]==d[b][(i+2)%6] && 
        d[a][3]==d[b][(i+3)%6] && 
        d[a][4]==d[b][(i+4)%6] && 
        d[a][5]==d[b][(i+5)%6]) || (

        d[a][0]==d[b][i] && 
        d[a][1]==d[b][(i+5)%6] && 
        d[a][2]==d[b][(i+4)%6] && 
        d[a][3]==d[b][(i+3)%6] && 
        d[a][4]==d[b][(i+2)%6] && 
        d[a][5]==d[b][(i+1)%6]))
        return true;
    }
    return false;
}
int main()
{
    int n;
    int i,j;
    scanf("%d",&n);
    for(i=0;i<n;++i)
    {
        int sum=0;
        for(j=0;j<6;++j)
        {
            scanf("%d",&d[i][j]);
            sum+=d[i][j];
        }
        sum%=MAXVALUE;
        for(j=0;j<hash[sum].size();++j)
        {
            if(isSame(hash[sum][j],i))
            {
                printf("%s\n","Twin snowflakes found.");
                exit(0);
            }
        }
        hash[sum].push_back(i);
    }
    printf("%s\n","No two snowflakes are alike.");
    return 0;
}