| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 40969 | Accepted: 10792 |
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.
题意：判断是否有两片相同的雪花，每个雪花有六片花瓣，顺时针或逆时针比较，如果每片长度都相同，则为两片相同的花瓣。由于如果雪花相同，每片雪花的所有花瓣长度之和取余得到的数一定相同，所以将每片雪花的所有花瓣长度之和拿来hash得到key，将雪花分类。
如果一类雪花不止一个则有可能有相同的雪花，再在该类中逐个进行判断。
1 #include<iostream> 2 #include<vector> 3 #include<cstdio> 4 5 using namespace std; 6 7 int n; 8 const int prime = 99991; 9 struct node 10 { 11 int num[6]; 12 }; 13 vector<node>ve[prime]; 14 15 void hashh(node a) 16 { 17 int key = (a.num[0]+a.num[1]+a.num[2]+a.num[3]+a.num[4]+a.num[5])%prime; 18 ve[key].push_back(a); 19 } 20 int main() 21 { 22 scanf("%d",&n); 23 node a; 24 for(int i = 0 ; i< n; i++) 25 { 26 for(int j = 0; j < 6; j++) 27 scanf("%d",&a.num[j]); 28 hashh(a); 29 } 30 31 for(int i = 0; i < prime; i++) 32 { 33 if(ve[i].size()>1) 34 { 35 for(int j = 0; j < ve[i].size()-1 ; j++) 36 { 37 for(int k = j+1; k < ve[i].size() ; k++) 38 { 39 for(int z = 0; z < 6; z++) 40 { 41 int pos=0; 42 if(ve[i][k].num[z]==ve[i][j].num[pos]) 43 { 44 int con=0; 45 for(int w = z+1; w < 6; w++) 46 if(ve[i][k].num[w]==ve[i][j].num[++pos]) 47 con++; 48 else 49 break; 50 for(int x = 0; x < z; x++) 51 if(ve[i][k].num[x]==ve[i][j].num[++pos]) 52 con++; 53 else 54 break; 55 if(con==5) 56 { 57 printf("Twin snowflakes found. "); 58 return 0; 59 } 60 con=0; 61 pos=0; 62 for(int w = z-1; w >= 0; w--) 63 if(ve[i][k].num[w]==ve[i][j].num[++pos]) 64 con++; 65 else 66 break; 67 for(int x = 5; x > z; x--) 68 if(ve[i][k].num[x]==ve[i][j].num[++pos]) 69 con++; 70 else 71 break; 72 if(con==5) 73 { 74 printf("Twin snowflakes found. "); 75 return 0; 76 } 77 } 78 } 79 80 } 81 } 82 } 83 } 84 printf("No two snowflakes are alike. "); 85 return 0; 86 }