Catching Fish
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1741 Accepted Submission(s): 686
Problem Description
Ignatius
likes catching fish very much. He has a fishnet whose shape is a circle
of radius one. Now he is about to use his fishnet to catch fish. All
the fish are in the lake, and we assume all the fish will not move when
Ignatius catching them. Now Ignatius wants to know how many fish he can
catch by using his fishnet once. We assume that the fish can be regard
as a point. So now the problem is how many points can be enclosed by a
circle of radius one.
Note: If a fish is just on the border of the fishnet, it is also caught by Ignatius.
Note: If a fish is just on the border of the fishnet, it is also caught by Ignatius.
Input
The
input contains several test cases. The first line of the input is a
single integer T which is the number of test cases. T test cases follow.
Each test case starts with a positive integer N(1<=N<=300) which indicate the number of fish in the lake. Then N lines follow. Each line contains two floating-point number X and Y (0.0<=X,Y<=10.0). You may assume no two fish will at the same point, and no two fish are closer than 0.0001, no two fish in a test case are approximately at a distance of 2.0. In other words, if the distance between the fish and the centre of the fishnet is smaller 1.0001, we say the fish is also caught.
Each test case starts with a positive integer N(1<=N<=300) which indicate the number of fish in the lake. Then N lines follow. Each line contains two floating-point number X and Y (0.0<=X,Y<=10.0). You may assume no two fish will at the same point, and no two fish are closer than 0.0001, no two fish in a test case are approximately at a distance of 2.0. In other words, if the distance between the fish and the centre of the fishnet is smaller 1.0001, we say the fish is also caught.
Output
For each test case, you should output the maximum number of fish Ignatius can catch by using his fishnet once.
Sample Input
4
3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
Sample Output
2
5
5
11
模板题:
#include<cstdio> #include<cstring> #include<algorithm> #include<math.h> using namespace std; const int N = 300; struct Point { double x,y; } p[N]; struct Node { double angle; bool in; } arc[180000]; int n,cnt; double R; double dist(Point p1,Point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } bool cmp(Node n1,Node n2) { return n1.angle!=n2.angle?n1.angle<n2.angle:n1.in>n2.in; } void MaxCircleCover() { int ans=1; for(int i=0; i<n; i++) { int cnt=0; for(int j=0; j<n; j++) { if(i==j) continue; if(dist(p[i],p[j])>R*2) continue; double angle=atan2(p[i].y-p[j].y,p[i].x-p[j].x); double phi=acos(dist(p[i],p[j])/2); arc[cnt].angle=angle-phi; arc[cnt++].in=true; arc[cnt].angle=angle+phi; arc[cnt++].in=false; } sort(arc,arc+cnt,cmp); int tmp=1; for(int i=0; i<cnt; i++) { if(arc[i].in) tmp++; else tmp--; ans=max(ans,tmp); } } printf("%d ",ans); } int main() { int tcase; scanf("%d",&tcase); while(tcase--) { scanf("%d",&n); //scanf("%lf",&R); R = 1; //此题 R 为 1 for(int i=0; i<n; i++) scanf("%lf%lf",&p[i].x,&p[i].y); MaxCircleCover(); } return 0; }