poj1981Circle and Points(单位圆覆盖最多的点）

链接

O（n^3）的做法：

枚举任意两点为弦的圆，然后再枚举其它点是否在圆内。

用到了两个函数

atan2反正切函数，据说可以很好的避免一些特殊情况

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<vector>
 7 #include<cmath>
 8 #include<queue>
 9 #include<set>
10 using namespace std;
11 #define N 310
12 #define LL long long
13 #define INF 0xfffffff
14 const double eps = 1e-8;
15 const double pi = acos(-1.0);
16 const double inf = ~0u>>2;
17 struct point
18 {
19     double x,y;
20     point(double x = 0,double y =0 ):x(x),y(y){}
21 }p[N];
22 double dis(point a,point b)
23 {
24     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
25 }
26 point getcircle(point p1,point p2)
27 {
28     point mid = point((p1.x+p2.x)/2,(p2.y+p1.y)/2);
29     double angle = atan2(p2.y-p1.y,p2.x-p1.x);
30     double d = sqrt(1.0-dis(p1,mid)*dis(p1,mid));
31     return point(mid.x+d*sin(angle),mid.y-d*cos(angle));
32 }
33 int dcmp(double x)
34 {
35     if(fabs(x)<eps)return 0;
36     else return x<0?-1:1;
37 }
38 int main()
39 {
40     int i,j,n;
41     while(scanf("%d",&n)&&n)
42     {
43         for(i = 1 ;i <= n; i++)
44         scanf("%lf%lf",&p[i].x,&p[i].y);
45         int maxz = 1;
46         for(i = 1; i <= n; i++)
47             for(j = i+1 ; j <= n ;j++)
48             {
49                 if(dis(p[i],p[j])>2.0) continue;
50                 int tmax = 0;
51                 point cir = getcircle(p[i],p[j]);
52                 for(int g =  1; g <= n ;g++)
53                 {
54                     if(dcmp(dis(cir,p[g])-1.0)>0)
55                     continue;
56                     tmax++;
57                 }
58                 maxz = max(maxz,tmax);
59             }
60         printf("%d\n",maxz);
61     }
62     return 0;
63 }

O（n^2log(n))

这个类似扫描线的做法，以每一个点为圆心化圆，枚举与其相交得圆，保存交点和角度，按角度排序后，扫一遍。

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<vector>
 7 #include<cmath>
 8 #include<queue>
 9 #include<set>
10 using namespace std;
11 #define N 310
12 #define LL long long
13 #define INF 0xfffffff
14 const double eps = 1e-8;
15 const double pi = acos(-1.0);
16 const double inf = ~0u>>2;
17 struct point
18 {
19     double x,y;
20     point(double x = 0,double y =0 ):x(x),y(y) {}
21 } p[N];
22 struct node
23 {
24     double ang;
25     int in;
26 } arc[N*N];
27 double dis(point a,point b)
28 {
29     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
30 }
31 int dcmp(double x)
32 {
33     if(fabs(x)<eps)return 0;
34     else return x<0?-1:1;
35 }
36 bool cmp(node a,node b)
37 {
38     if(dcmp(a.ang-b.ang)==0)
39         return a.in>b.in;
40     return dcmp(a.ang-b.ang)<0;
41 }
42 int main()
43 {
44     int i,j,n;
45     while(scanf("%d",&n)&&n)
46     {
47         for(i = 1 ; i <= n; i++)
48             scanf("%lf%lf",&p[i].x,&p[i].y);
49         int g = 0;
50         int ans = 0,maxz = 1;
51         for(i = 1; i <= n ; i++)
52         {
53             ans = 0;
54             g = 0;
55             for(j = 1; j <= n ; j++)
56             {
57                 if(dis(p[i],p[j])>2.0) continue;
58                 double ang1 = atan2(p[j].y-p[i].y,p[j].x-p[i].x);
59                 double ang2 = acos(dis(p[i],p[j])/2);
60                 arc[++g].ang = ang1-ang2;//这里角度的算法很巧妙
61                 arc[g].in = 1;
62                 arc[++g].ang = ang1+ang2;
63                 arc[g].in = -1;
64             }
65             sort(arc+1,arc+g+1,cmp);
66
67             //cout<<g<<endl;
68             for(j = 1 ; j <= g;j++)
69             {
70                 ans+=arc[j].in;
71                 maxz = max(maxz,ans);
72             }
73         }
74         printf("%d\n",maxz);
75     }
76     return 0;
77 }

poj1981Circle and Points(单位圆覆盖最多的点）