Time Limit: 5000MS | Memory Limit: 30000K | |
Total Submissions: 7327 | Accepted: 2651 | |
Case Time Limit: 2000MS |
Description
Fig 1. Circle and Points
Input
You may assume 1 <= N <= 300, 0.0 <= X <= 10.0, and 0.0 <= Y <= 10.0. No two points are closer than 0.0001. No two points in a data set are approximately at a distance of 2.0. More precisely, for any two points in a data set, the distance d between the two never satisfies 1.9999 <= d <= 2.0001. Finally, no three points in a data set are simultaneously very close to a single circle of radius one. More precisely, let P1, P2, and P3 be any three points in a data set, and d1, d2, and d3 the distances from an arbitrarily selected point in the xy-plane to each of them respectively. Then it never simultaneously holds that 0.9999 <= di <= 1.0001 (i = 1, 2, 3).
Output
Sample Input
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 0
Sample Output
2 5 5 11
代码转自,不想去弄了。。以后就做模板用好了 http://www.cnblogs.com/-sunshine/archive/2012/10/11/2719859.html
贴个模板:
#include<stdio.h> #include<iostream> #include<string.h> #include <stdlib.h> #include<math.h> #include<algorithm> 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\n",ans); } int main(){ while(scanf("%d",&n)!=EOF&&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; }