HDU-1007-最小公共点对
http://acm.hdu.edu.cn/showproblem.php?pid=1007
Quoit Design
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58813 Accepted Submission(s): 15582
Problem Description
Have
you ever played quoit in a playground? Quoit is a game in which flat
rings are pitched at some toys, with all the toys encircled awarded.
In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring.
Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0.
In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring.
Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0.
Input
The
input consists of several test cases. For each case, the first line
contains an integer N (2 <= N <= 100,000), the total number of
toys in the field. Then N lines follow, each contains a pair of (x, y)
which are the coordinates of a toy. The input is terminated by N = 0.
Output
For
each test case, print in one line the radius of the ring required by
the Cyberground manager, accurate up to 2 decimal places.
Sample Input
2
0 0
1 1
2
1 1
1 1
3
-1.5 0
0 0
0 1.5
0
Sample Output
0.71
0.00
0.75
Author
CHEN, Yue
Source
给出n个点，找到一个最大半径的圆，满足这个圆每次最多只能覆盖一个点。输出这个最大圆的半径。显然这个最大半径就是最近公共点对的一半，二分找最近距离就好了。、
1 #include<iostream> 2 #include<cmath> 3 #include<cstdio> 4 #include<algorithm> 5 #include<cstring> 6 #include<vector> 7 using namespace std; 8 #define inf 0x3f3f3f3f 9 struct Point 10 { 11 double x,y; 12 }P[100010]; 13 inline bool cmpx(Point A,Point B){return A.x<B.x;} 14 inline bool cmpy(Point A,Point B){return A.y<B.y;} 15 double dis(Point A,Point B) 16 { 17 double dx=(A.x-B.x)*(A.x-B.x); 18 double dy=(A.y-B.y)*(A.y-B.y); 19 return sqrt(dx+dy); 20 } 21 double solve(int l,int r) 22 { 23 if(l==r) return inf; 24 if(l+1==r) return dis(P[l],P[r]); 25 vector<Point>vp; 26 int mid=(l+r)>>1; 27 double res=min(solve(l,mid),solve(mid+1,r)); 28 for(int i=l;i<=r;++i) 29 if(P[i].x>=P[mid].x-res&&P[i].x<=P[mid].x+res) 30 vp.push_back(P[i]); 31 sort(vp.begin(),vp.end(),cmpy); 32 for(int i=0;i<vp.size();++i){ 33 for(int j=1;i+j<vp.size()&&j<7;++j){ 34 if(res>dis(vp[i],vp[i+j])) 35 res=dis(vp[i],vp[i+j]); 36 } 37 } 38 return res; 39 } 40 int main() 41 { 42 int n; 43 while(cin>>n&&n){ 44 for(int i=1;i<=n;++i) 45 scanf("%lf%lf",&P[i].x,&P[i].y); 46 sort(P+1,P+1+n,cmpx); 47 printf("%.2f\n",solve(1,n)/2.0); 48 } 49 return 0; 50 }
