題目大意是找一個最小的圓覆蓋給你一堆點中的兩個點,輸出改圓的半徑
基礎最近點對模板題
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#define LL __int64
using namespace std;
const int maxn = 100010;
struct point
{
double x,y;
};
point p1[maxn],p2[maxn];
int n;
bool cmpx(point a,point b)//按x升序
{
return a.x < b.x;
}
bool cmpy(point a,point b)//按y降序
{
return a.y < b.y;
}
double dis(point x,point y)
{
double v = (x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y);
return sqrt(v);
}
double solve(int l,int r)
{
if(l+1==r) return dis(p1[l],p1[r]);//這裏有點像搜索,如果到了最近幾個數,我就直接比較輸出
if(l+2==r) return min(dis(p1[l],p1[l+1]),min(dis(p1[l+1],p1[l+2]),dis(p1[l],p1[r])));
int mid = (l+r)>>1;//二分思想
double ans = min(solve(l,mid),solve(mid+1,r));
int cnt = 0;
for(int i=l;i<=r;i++){ //在這個區域找點
if(fabs(p1[i].x-p1[mid].x)<=ans)
p2[cnt++] = p1[i];
}
sort(p2,p2+cnt,cmpy);
for(int i=0;i<cnt;i++){
for(int j=i+1;j<cnt;j++){
if(p2[j].y - p2[i].y >= ans) break;//優化判斷
ans = min(ans,dis(p2[i],p2[j]));
}
}
return ans;
}
int main()
{
while(scanf("%d",&n)&&n){
for(int i=0;i<n;i++) scanf("%lf%lf",&p1[i].x,&p1[i].y);
sort(p1,p1+n,cmpx);
double ans = solve(0,n-1);//找出直徑
printf("%.2f\n",ans/2.0);//輸出半徑
}
return 0;
}