Snowflake Snow Snowflakes - POJ 3349 哈希

最新推荐文章于 2020-09-06 17:13:10 发布

提比-我有特殊的AC技巧

最新推荐文章于 2020-09-06 17:13:10 发布

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本文介绍了一个编程问题，即如何通过比较六边形雪花各臂的长度来判断是否存在完全相同的两朵雪花。通过哈希和比较算法，实现了快速查找可能相同的雪花。

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Snowflake Snow Snowflakes

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 29416		Accepted: 7748

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

题意：判断有无两个相同的雪花，注意比较的时候也有顺序。

思路：哈希得到哈希值相同的，就好比较了。PS：看网上有直接把哈希值相同的雪花先sort下，也能AC的，但是1 2 3 4 5 6 和 1 3 2 4 5 6其实不是一样的，测试数据有点水吧。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int snow[100010][6];
vector<int> hash[5000];
bool flag=false;
bool solve(int a,int b)
{ for(int i=0;i<=5;i++)
  { if(( snow[a][0]==snow[b][i] &&
        snow[a][1]==snow[b][(i+1)%6] &&
        snow[a][2]==snow[b][(i+2)%6] &&
        snow[a][3]==snow[b][(i+3)%6] &&
        snow[a][4]==snow[b][(i+4)%6] &&
        snow[a][5]==snow[b][(i+5)%6]
      )||
      ( snow[a][5]==snow[b][i] &&
        snow[a][4]==snow[b][(i+1)%6] &&
        snow[a][3]==snow[b][(i+2)%6] &&
        snow[a][2]==snow[b][(i+3)%6] &&
        snow[a][1]==snow[b][(i+4)%6] &&
        snow[a][0]==snow[b][(i+5)%6]
      ))
      {flag=true;return true;}
  }
  return false;
}
int main()
{ int n,i,j,k,sum;
  scanf("%d",&n);
  for(k=1;k<=n;k++)
  { sum=0;
    for(i=0;i<=5;i++)
    { scanf("%d",&snow[k][i]);
      sum+=snow[k][i];
    }
    sum%=5000;
    for(i=0;i<hash[sum].size();i++)
     solve(k,hash[sum][i]);
    if(flag)
     break;
    hash[sum].push_back(k);
  }
  if(flag)
   printf("Twin snowflakes found.\n");
  else
   printf("No two snowflakes are alike.\n");
}