标签：least cst name using tput arch ++ include cloc
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.
Hash不错的例题，详情见代码注释。
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #define P 99991//小于N的最大质数作为模数比较合适，这样Hash值重复的概率比较小 #define N 100005 using namespace std; int n; int a[10],tot=0,head[N],next[N],snow[N][6];//暂时存放每片雪花 int H(int *a)//每片雪花各个数的和与各个数的积之和作为哈希值比较高效率 { int sum=0,mul=1; int i; for(i=0;i<6;i++) { sum=(sum+a[i])%P; mul=(long long)mul*a[i]%P; } return (sum+mul)%P;//别忘了模一下 } bool equal(int *a,int *b) { int i,j,k; for(i=0;i<6;i++)//a从哪一位开始比 { for(j=0;j<6;j++)//b从哪一位开始比 { bool eq=1; for(k=0;k<6;k++)//a正着 b正着 走k个数 { if(a[(i+k)%6]!=b[(j+k)%6])eq=0; } if(eq)return 1; eq=1; for(k=0;k<6;k++)//a正着 b反着 走k个数 { if(a[(i+k)%6]!=b[(j-k+6)%6])eq=0;//为防止出现负数要加上模数 } if(eq)return 1; } } return 0;//都不行的话return 0 } bool add(int *a) { int val=H(a);//获得Hash值 int i; for(i=head[val];i;i=next[i])//类比邻接表进行遍历 { if(equal(snow[i],a))return 1;//如果相等直接返回1 } //没找到的话直接插入 ++tot;//类比邻接表的add操作 memcpy(snow[tot],a,6*sizeof(int));//把当前新雪花插入表 next[tot]=head[val];//回忆链表操作 head[val]=tot; return 0; } int main() { cin>>n; int i; for(i=1;i<=n;i++)//先从主函数读是一个好习惯 { int j; for(j=0;j<6;j++) { scanf("%d",&a[j]); } if(add(a)) { cout<<"Twin snowflakes found."; return 0; } } cout<<"No two snowflakes are alike."; return 0; }
POJ3349 Snowflake Snow Snowflakes（Hash）
标签：least cst name using tput arch ++ include cloc
原文地址：https://www.cnblogs.com/lipoicyclic/p/12623897.html
