POJ 1279 Art Gallery(半平面交)

题目链接

回忆了一下，半平面交，整理了一下模版。

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <string>

 4 #include <cmath>

 5 #include <algorithm>

 6 using namespace std;

 7 #define eps 1e-8

 8 #define N 2001

 9 struct point

10 {

11     double x,y;

12 }p[N],temp[N],pre[N];

13 int n,m;

14 double a,b,c;

15 void getline(point x,point y)

16 {

17     a = y.y - x.y;

18     b = x.x - y.x;

19     c = y.x * x.y - x.x * y.y;

20 }

21 point intersect(point x,point y) //获取直线ax+by+c==0  和点x和y所连直线的交点

22 {

23     double u = fabs(a*x.x+b*x.y+c);

24     double v = fabs(a*y.x+b*y.y+c);

25     point ans;

26     ans.x = (x.x*v+y.x*u)/(u+v);

27     ans.y = (x.y*v+y.y*u)/(u+v);

28     return ans;

29 }

30 void cut()//坐标顺时针给出，注释为逆时针

31 {

32     int num = 0,i;

33     for(i = 1;i <= m;i ++)

34     {

35         if(a*p[i].x + b*p[i].y + c >= -eps)//<= eps

36         {

37             temp[++num] = p[i];

38         }

39         else

40         {

41             if(a*p[i-1].x + b*p[i-1].y + c > eps)//< -eps

42             temp[++num] = intersect(p[i],p[i-1]);

43             if(a*p[i+1].x + b*p[i+1].y + c > eps)

44             temp[++num] = intersect(p[i],p[i+1]);//< -eps

45         }

46     }

47     for(i = 1;i <= num;i ++)

48     p[i] = temp[i];

49     p[0] = p[num];

50     p[num+1] = p[1];

51     m = num;

52 }

53 void fun()

54 {

55     int i;

56     m = n;

57     for(i = 1;i <= n;i ++)

58     {

59         getline(pre[i],pre[i+1]);

60         cut();

61     }

62 }

63 int main()

64 {

65     int i,t;

66     scanf("%d",&t);

67     while(t--)

68     {

69         scanf("%d",&n);

70         for(i = 1;i <= n;i ++)

71         {

72             scanf("%lf%lf",&pre[i].x,&pre[i].y);

73             p[i] = pre[i];

74         }

75         pre[n+1] = pre[1];

76         p[n+1] = p[1];

77         p[0] = p[n];

78         fun();

79         double s1,s2;

80         s1 = s2 = 0;

81         for(i = 0;i < m;i ++)

82         {

83             s1 += p[(i+1)%n].y*p[i].x;

84             s2 += p[(i+1)%n].x*p[i].y;

85         }

86         printf("%.2f\n",fabs(s2-s1)/2+eps);

87     }

88     return 0;

89 }