Enum:Game of Lines(POJ 3668)

　　　　　　　　　　　　　　

　　　　　　　　　　　　　　　　　　画直线

　　题目大意：给定一些点集，要你找两点之间的连线不平行的有多少条

　　数据量比较少，直接暴力枚举，然后放到set查找即可

 #include <iostream>
 #include <functional>
 #include <algorithm>
 #include <set>

 using namespace std;

 static struct _p_set
 {
     long double x, y;
 }points[];
 set<long double>lines;

 int gcd(const int, const int);

 int main(void)
 {
     int point_sum, cut;
     long double tmp;
     //pair<int,int>tmp;
     while (~scanf("%d", &point_sum))
     {
         lines.clear();
         for (int i = ; i < point_sum; i++)
             scanf("%lf%lf", &points[i].x, &points[i].y);
         for (int i = ; i < point_sum; i++)
         {
             for (int j = i + ; j < point_sum; j++)
             {
                 if ((points[i].x - points[j].x) != )
                     tmp = (points[i].y - points[j].y) / (points[i].x - points[j].x);
                 else
                     tmp = (long double)INT_MAX;
                 lines.insert(tmp);
             }
         }
         printf("%d\n", lines.size());
     }

     return EXIT_SUCCESS;
 }

　　

　　在讨论版那里还找到了一种很新奇的做法，可以无视除数是0和精度的问题

 #include <iostream>
 #include <functional>
 #include <algorithm>
 #include <set>

 using namespace std;

 static struct _p_set
 {
     int x, y;
 }points[];
 set<pair<int,int>>lines;

 int gcd(const int, const int);

 int main(void)
 {
     int point_sum, cut;
     pair<int,int>tmp;
     while (~scanf("%d", &point_sum))
     {
         lines.clear();
         for (int i = ; i < point_sum; i++)
             scanf("%d%d", &points[i].x, &points[i].y);
         for (int i = ; i < point_sum; i++)
         {
             for (int j = i + ; j < point_sum; j++)
             {
                 cut = gcd(points[i].y - points[j].y, points[i].x - points[j].x);
                 tmp.first = (points[i].y - points[j].y) / cut;
                 tmp.second = (points[i].x - points[j].x) / cut;
                 lines.insert(tmp);
             }
         }
         printf("%d\n", lines.size());
     }
     return EXIT_SUCCESS;
 }

 int gcd(const int a, const int b)
 {
     if (b == )
         return a;
     return gcd(b, a%b);
 }

　　

　　其实时间差不多