LA 4064 (计数 极角排序) Magnetic Train Tracks

这个题和UVa11529很相似。

枚举一个中心点，然后按极角排序，统计以这个点为钝角的三角形的个数，然后用C(n, 3)减去就是答案。

另外遇到直角三角形的情况很是蛋疼，可以用一个eps，不嫌麻烦的话就用整数的向量做点积。

 1 #include <cstdio>

 2 #include <cmath>

 3 #include <algorithm>

 4 using namespace std;

 5 

 6 typedef long long LL;

 7 const int maxn = 1200 + 10;

 8 const double PI = acos(-1.0);

 9 const double eps = 1e-9;

10 

11 struct Point

12 {

13     double x, y;

14     Point() {}

15     Point(double x, double y):x(x), y(y) {}

16 }p[maxn], p2[maxn];

17 

18 Point operator - (const Point& A, const Point& B)

19 {

20     return Point(A.x-B.x, A.y-B.y);

21 }

22 

23 double ang[maxn * 2];

24 

25 LL inline C3(int n) { return (LL)n * (n-1) / 2 * (n-2) / 3; }

26 

27 int main()

28 {

29     //freopen("in.txt", "r", stdin);

30     int n, kase = 0;

31 

32     while(scanf("%d", &n) == 1 && n)

33     {

34         for(int i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);

35 

36         LL cnt = 0;

37         for(int i = 0; i < n; i++)

38         {

39             int k = 0;

40             for(int j = 0; j < n; j++) if(j != i)

41             {

42                 p2[k] = p[j] - p[i];

43                 ang[k] = atan2(p2[k].y, p2[k].x);

44                 ang[k + n - 1] = ang[k] + PI * 2.0;

45                 k++;

46             }

47             k = 2*n-2;

48             sort(ang, ang + k);

49             int L, R1 = 0, R2 = 0;

50             for(L = 0; L < n-1; L++)

51             {

52                 double b1 = ang[L] + PI / 2;

53                 double b2 = ang[L] + PI;

54                 while(ang[R1] <= b1 - eps) R1++;

55                 while(ang[R2] < b2) R2++;

56                 cnt += R2 - R1;

57             }

58         }

59         LL ans = C3(n) - cnt;

60         printf("Scenario %d:\nThere are %lld sites for making valid tracks\n", ++kase, ans);

61     }

62 

63     return 0;

64 }

代码君