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

 1 #include <iostream>
 2 #include <functional>
 3 #include <algorithm>
 4 #include <set>
 5 
 6 using namespace std;
 7 
 8 static struct _p_set
 9 {
10     long double x, y;
11 }points[201];
12 set<long double>lines;
13 
14 int gcd(const int, const int);
15 
16 int main(void)
17 {
18     int point_sum, cut;
19     long double tmp;
20     //pair<int,int>tmp;
21     while (~scanf("%d", &point_sum))
22     {
23         lines.clear();
24         for (int i = 0; i < point_sum; i++)
25             scanf("%lf%lf", &points[i].x, &points[i].y);
26         for (int i = 0; i < point_sum; i++)
27         {
28             for (int j = i + 1; j < point_sum; j++)
29             {
30                 if ((points[i].x - points[j].x) != 0)
31                     tmp = (points[i].y - points[j].y) / (points[i].x - points[j].x);
32                 else
33                     tmp = (long double)INT_MAX;
34                 lines.insert(tmp);
35             }
36         }
37         printf("%d
", lines.size());
38     }
39 
40     return EXIT_SUCCESS;
41 }

　　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

 1 #include <iostream>
 2 #include <functional>
 3 #include <algorithm>
 4 #include <set>
 5 
 6 using namespace std;
 7 
 8 static struct _p_set
 9 {
10     int x, y;
11 }points[201];
12 set<pair<int,int>>lines;
13 
14 int gcd(const int, const int);
15 
16 int main(void)
17 {
18     int point_sum, cut;
19     pair<int,int>tmp;
20     while (~scanf("%d", &point_sum))
21     {
22         lines.clear();
23         for (int i = 0; i < point_sum; i++)
24             scanf("%d%d", &points[i].x, &points[i].y);
25         for (int i = 0; i < point_sum; i++)
26         {
27             for (int j = i + 1; j < point_sum; j++)
28             {
29                 cut = gcd(points[i].y - points[j].y, points[i].x - points[j].x);
30                 tmp.first = (points[i].y - points[j].y) / cut;
31                 tmp.second = (points[i].x - points[j].x) / cut;
32                 lines.insert(tmp);
33             }
34         }
35         printf("%d
", lines.size());
36     }
37     return EXIT_SUCCESS;
38 }
39 
40 int gcd(const int a, const int b)
41 {
42     if (b == 0)
43         return a;
44     return gcd(b, a%b);
45 }

　　其实时间差不多