poj 3349 Snowflake Snow Snowflakes
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Snowflake Snow Snowflakes
Time Limit: 4000MS Memory Limit: 65536K
Total Submissions: 44089 Accepted: 11538
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
附上题目链接:http://poj.org/problem?id=3349;
思路:(注释里写的解决了我当时的疑惑)
两个雪花是否相等,我们可以枚举,顺逆比较。
O(n2)比较?
所以我们要用hash的方法做这道题。
如果两片雪花相同,那么它们六个角的和必然相同。
所以我们把六个角和相同的hash到一个容器里。
把同一个容器里的雪花拿出来比较。
这样大大缩减了时间。
代码:
#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;const int N=100005;const int prime=100000;int n;int snow[N][7];vector<int >V[prime];int cmp(int pos1, int pos2){ int i, j; for(i=0;i<6;i++) { if(snow[pos1][0]==snow[pos2][i]) { for(j=1;j<6;j++) if(snow[pos1][j]!=snow[pos2][(i+j)%6]) //顺时针; break; if(j==6) return true; for(j=1;j<6;j++) if(snow[pos1][6-j]!=snow[pos2][(i+j)%6]) //逆时针; break; if(j==6) return true; } } return false;}int check(){ for(int i=0;i<prime;i++) //依次在每个边数相等的vector[i]里找是否存在两个相同的雪花; for(int j=0;j<V[i].size();j++) for(int k=j+1;k<V[i].size();k++) if(cmp(V[i][j], V[i][k])) return 1; return 0;}int main(){ while(~scanf("%d", &n)) { for(int i=0;i<n;i++) { int sum=0; for(int j=0;j<6;j++) { scanf("%d", &snow[i][j]); sum+=snow[i][j]; } V[sum%prime].push_back(i); //把那些6个边相加之和相等的雪花存到一个vector[i]里; } if(check()) printf("Twin snowflakes found.\n"); else printf("No two snowflakes are alike.\n"); }}//这个题的key是雪花边数相加之和;
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