计算几何的几道题

叉积求面积

hdu2036（模版题）:

http://acm.hdu.edu.cn/showproblem.php?pid=2036

View Code

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 #include<math.h>

 5 using namespace std;

 6 struct node

 7 {

 8     int x,y;

 9 }q[101];

10 double cross(node a,node b)

11 {

12     return a.x*b.y-a.y*b.x;

13 }

14 int main()

15 {

16     int i,j,k,n;

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

18     {

19         double s = 0;

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

21         scanf("%d%d",&q[i].x,&q[i].y);

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

23         {

24             if(i<n)

25             s+=cross(q[i],q[i+1]);

26             else

27             s+=cross(q[n],q[1]);

28         }

29         if(s<0)

30         s  = -s;

31         s = s/2;

32         printf("%.1lf\n",s);

33     }

34     return 0;

35 }

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=41

求怎样旋转这个半圆能包围最多的点

 

1. 到圆心的距离大于半径的点直接排除。

  

2. 以圆心和任意一点确定一 有向线段作为半径位置，分别计数该有向线段左边点的个数 (nl) 和右边点的个数 (nr) 。

 

重复步骤 2 直到所有点都被枚举

过。

4. 枚举过程中出现的最大的 nl 或

nr 就是所求的结果。

View Code

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 using namespace std;

 5 struct node

 6 {

 7     int x,y;

 8 }no[200];

 9 int cross(int x1,int y1,int x2,int y2)

10 {

11     return x1*y2-x2*y1;

12 }

13 int main()

14 {

15     int i,j,k,n,m,x,y,g;

16     double r;

17     while(scanf("%d%d%lf",&n,&m,&r)!=EOF)

18     {

19         if(r<0)

20         break;

21         scanf("%d",&k);

22         g = 0;

23         for(i = 1; i <= k ;i++)

24         {

25             scanf("%d%d",&x,&y);

26             if((x-n)*(x-n)+(y-m)*(y-m)<=r*r)//排除到圆心距离大于半径的点

27             {

28                 g++;

29                 no[g].x = x;

30                 no[g].y = y;

31             }

32         }

33         int max = 0;

34         for(i = 1; i <= g ; i++)

35         {

36             int numl = 0,numr = 0;

37             for(j = 1; j <= g ; j++)

38             {

39                 int d = cross(no[i].x-n,no[i].y-m,no[j].x-n,no[j].y-m);//点乘判断是哪边的点

40                 if(d==0)

41                 {

42                     numl++;

43                     numr++;

44                 }

45                 if(d>0)

46                 numl++;

47                 if(d<0)

48                 numr++;

49             }

50             if(numl>max)

51             max = numl;

52             if(numr>max)

53             max = numr;

54         }

55         printf("%d\n",max);

56     }

57     return 0;

58 }

http://acm.hdu.edu.cn/showproblem.php?pid=1086

规范相交题目 利用差乘判断相交 不用考虑重点问题

View Code

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 using namespace std;

 5 double eps=0.0000000001;

 6 struct node

 7 {

 8     double x1,x2,y1,y2;

 9 };

10 double cross(double x1,double x2,double y1,double y2)

11 {

12     return x1*y2-x2*y1;

13 }

14 int main()

15 {

16     int i,j,k,n,m;

17     node q[101];

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

19     {

20         int num = 0;

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

22         {

23             scanf("%lf%lf%lf%lf",&q[i].x1,&q[i].y1,&q[i].x2,&q[i].y2);

24         }

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

26         for(j = i+1 ; j <= n ; j++)

27         {

28             double d1 = cross(q[i].x1-q[i].x2,q[i].y1-q[i].y2,q[i].x1-q[j].x1,q[i].y1-q[j].y1);

29             double d2 = cross(q[i].x1-q[i].x2,q[i].y1-q[i].y2,q[i].x1-q[j].x2,q[i].y1-q[j].y2);

30             double d3 = cross(q[j].x1-q[j].x2,q[j].y1-q[j].y2,q[j].x1-q[i].x2,q[j].y1-q[i].y2);

31             double d4 = cross(q[j].x1-q[j].x2,q[j].y1-q[j].y2,q[j].x1-q[i].x1,q[j].y1-q[i].y1);

32             if(d1*d2<eps&&d3*d4<eps)

33             num++;

34         }

35         printf("%d\n",num);

36     }

37     return 0;

38 }

  http://poj.org/problem?id=2318

二分查找+点乘判断左右

View Code

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 using namespace std;

 5 struct node

 6 {

 7     int x1,x2;

 8 }q[5001];

 9 int cross(node a,node b)

10 {

11     if(a.x1*b.x2-a.x2*b.x1>0)

12     return 1;

13     else

14     return 0;

15 }

16 int x1,x2,n,y1,y2;

17 int judge(int x,int y)

18 {

19     int l =0,r = n+1,mid,k;

20     node a,b;

21     while(l<r)

22     {

23         mid = (l+r)/2;

24         if(mid==l)

25         return mid;

26         a.x1 = q[mid].x1-q[mid].x2;

27         a.x2 = y1-y2;

28         b.x1 = x-q[mid].x2;

29         b.x2 = y-y2;

30         if(cross(a,b))

31         r = mid;

32         else

33         l = mid;

34     }

35     return mid;

36 }

37 int num[5010];

38 int main()

39 {

40     int i,j,m,t,x,y;

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

42     {

43         memset(num,0,sizeof(num));

44         scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);

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

46         scanf("%d%d",&q[i].x1,&q[i].x2);

47         while(m--)

48         {

49             scanf("%d%d",&x,&y);

50             int k = judge(x,y);

51             num[k]++;

52         }

53         for(i = 0 ; i <= n ; i++)

54         printf("%d: %d\n",i,num[i]);

55         puts("");

56     }

57     return 0;

58 }