poj 2653（线段相交）

第一道计算几何题。。。。

试用了下两种删除vector中元素的方法。

第一种使用：（每次erase一个元素后都会指向下一个元素位置）

1 for(int j=0;j<v.size();)

2 {

3     if(intersect_in(lin[i],lin[v[j]]))

4         v.erase(v.begin()+j);

5     else

6         j++;

7 }

View Code

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <vector>

 7 

 8 using namespace std;

 9 

10 #define eps 1e-8

11 #define PI acos(-1.0)//3.14159265358979323846

12 //判断一个数是否为0,是则返回true,否则返回false

13 #define zero(x)(((x)>0?(x):-(x))<eps)

14 //返回一个数的符号，正数返回1，负数返回2，否则返回0

15 #define _sign(x)((x)>eps?1:((x)<-eps?2:0))

16 

17 const int maxn =  1e5+5;

18 struct point 

19 {

20     double x,y;

21 };

22 struct line

23 {

24     point a,b;

25     line(){}

26     line(double x1,double y1,double x2,double y2)

27     {

28         a.x=x1;

29         a.y=y1;

30         b.x=x2;

31         b.y=y2;

32     }

33 };

34 line lin[maxn];

35 double xmult(point p1,point p2,point p0)

36 {

37     return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);

38 }

39 //计算dotproduct(P1-P0).(P2-P0)

40 double dmult(point p1,point p2,point p0)

41 {

42     return(p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y);

43 }

44 //判三点共线

45 int dots_inline(point p1,point p2,point p3)

46 {

47     return zero(xmult(p1,p2,p3));

48 }

49 //判点是否在线段上,包括端点

50 int dot_online_in(point p,line l)

51 {

52     return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps;

53 }

54 //判两点在线段同侧,点在线段上返回0

55 int same_side(point p1,point p2,line l)

56 {

57     return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps;

58 }

59 

60 int intersect_in(line u,line v)

61 {

62     if(!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))//规范相交

63         return!same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u);

64     return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u);

65 }

66 

67 int main()

68 {

69     int n;

70     double x1,x2,y1,y2;

71     while(scanf("%d",&n))

72     {

73         if(!n) break;

74         vector<int>v;

75         for(int i=1;i<=n;i++)

76         {

77             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);

78             lin[i]=line(x1,y1,x2,y2);

79             for(int j=0;j<v.size();)

80             {

81                 if(intersect_in(lin[i],lin[v[j]]))

82                     v.erase(v.begin()+j);

83                 else

84                     j++;

85             }

86             v.push_back(i);

87         }

88         printf("Top sticks:");

89         for(int i=0;i<v.size()-1;i++)

90             printf(" %d,",v[i]);

91         printf(" %d.\n",v[v.size()-1]);

92     }

93     return 0;

94 }

第二种使用remove_if()。

 1 line tmp;

 2 bool ok(line l)

 3 {

 4     if(intersect_in(tmp,l))

 5         return true;

 6     else

 7         return false;

 8 }

 9 

10 v.erase(remove_if(v.begin(),v.end(),ok),v.end());

View Code

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <cmath>

  5 #include <algorithm>

  6 #include <vector>

  7 

  8 using namespace std;

  9 

 10 #define eps 1e-8

 11 #define PI acos(-1.0)//3.14159265358979323846

 12 //判断一个数是否为0,是则返回true,否则返回false

 13 #define zero(x)(((x)>0?(x):-(x))<eps)

 14 //返回一个数的符号，正数返回1，负数返回2，否则返回0

 15 #define _sign(x)((x)>eps?1:((x)<-eps?2:0))

 16 

 17 struct point 

 18 {

 19     double x,y;

 20 };

 21 struct line

 22 {

 23     point a,b;

 24     int id;

 25     line(){}

 26     line(double x1,double y1,double x2,double y2,int i)

 27     {

 28         a.x=x1;

 29         a.y=y1;

 30         b.x=x2;

 31         b.y=y2;

 32         id=i;

 33     }

 34 };

 35 double xmult(point p1,point p2,point p0)

 36 {

 37     return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);

 38 }

 39 //计算dotproduct(P1-P0).(P2-P0)

 40 double dmult(point p1,point p2,point p0)

 41 {

 42     return(p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y);

 43 }

 44 //判三点共线

 45 int dots_inline(point p1,point p2,point p3)

 46 {

 47     return zero(xmult(p1,p2,p3));

 48 }

 49 //判点是否在线段上,包括端点

 50 int dot_online_in(point p,line l)

 51 {

 52     return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps;

 53 }

 54 //判两点在线段同侧,点在线段上返回0

 55 int same_side(point p1,point p2,line l)

 56 {

 57     return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps;

 58 }

 59 

 60 int intersect_in(line u,line v)

 61 {

 62     if(!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))//规范相交

 63         return!same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u);

 64     return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u);

 65 }

 66 

 67 line tmp;

 68 bool ok(line l)

 69 {

 70     if(intersect_in(tmp,l))

 71         return true;

 72     else

 73         return false;

 74 }

 75 

 76 int main()

 77 {

 78     int n;

 79     double x1,x2,y1,y2;

 80     while(scanf("%d",&n))

 81     {

 82         if(!n) break;

 83         vector<line>v;

 84         for(int i=1;i<=n;i++)

 85         {

 86             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);

 87             tmp=line(x1,y1,x2,y2,i);

 88             v.erase(remove_if(v.begin(),v.end(),ok),v.end());

 89             /*for(int j=0;j<v.size();)

 90             {

 91                 if(intersect_in(lin[i],lin[v[j]]))

 92                     v.erase(v.begin()+j);

 93                 else

 94                     j++;

 95             }*/

 96             v.push_back(tmp);

 97             

 98         }

 99         printf("Top sticks:");

100         for(int i=0;i<v.size()-1;i++)

101             printf(" %d,",v[i].id);

102         printf(" %d.\n",v[v.size()-1].id);

103     }

104     return 0;

105 }