hdu 1007 N个点中输出2点的最小距离的一半
分治法
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
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <string> 6 # include <cmath> 7 # include <queue> 8 # include <list> 9 # define LL long long 10 using namespace std ; 11 12 const double eps = 1e-6; 13 const int MAXN = 100010; 14 const double INF = 1e20; 15 struct Point 16 { 17 double x,y; 18 }; 19 20 double dist(Point a,Point b) 21 { 22 return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); 23 } 24 25 Point p[MAXN]; 26 Point tmpt[MAXN]; 27 bool cmpxy(Point a,Point b) 28 { 29 if(a.x != b.x)return a.x < b.x; 30 else return a.y < b.y; 31 } 32 bool cmpy(Point a,Point b) 33 { 34 return a.y < b.y; 35 } 36 double Closest_Pair(int left,int right) 37 { 38 double d = INF; 39 if(left == right)return d; 40 if(left + 1 == right) 41 return dist(p[left],p[right]); 42 int mid = (left+right)/2; 43 double d1 = Closest_Pair(left,mid); 44 double d2 = Closest_Pair(mid+1,right); 45 d = min(d1,d2); 46 int k = 0; 47 for(int i = left;i <= right;i++) 48 { 49 if(fabs(p[mid].x - p[i].x) <= d) 50 tmpt[k++] = p[i]; 51 } 52 sort(tmpt,tmpt+k,cmpy); 53 for(int i = 0;i <k;i++) 54 { 55 for(int j = i+1;j < k && tmpt[j].y - tmpt[i].y < d;j++) 56 { 57 d = min(d,dist(tmpt[i],tmpt[j])); 58 } 59 } 60 return d; 61 } 62 63 int main() 64 { 65 //freopen("in.txt","r",stdin) ; 66 int n; 67 while(scanf("%d",&n)==1 && n) 68 { 69 for(int i = 0;i < n;i++) 70 scanf("%lf%lf",&p[i].x,&p[i].y); 71 sort(p,p+n,cmpxy); 72 printf("%.2lf\n",Closest_Pair(0,n-1)/2); 73 } 74 return 0; 75 }
