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

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <string>
# include <cmath>
# include <queue>
# include <list>
# define LL long long
using namespace std ;

const double eps = 1e-6;
const int MAXN = 100010;
const double INF = 1e20;
struct Point
{
double x,y;
};

double dist(Point a,Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

Point p[MAXN];
Point tmpt[MAXN];
bool cmpxy(Point a,Point b)
{
if(a.x != b.x)return a.x < b.x;
else return a.y < b.y;
}
bool cmpy(Point a,Point b)
{
return a.y < b.y;
}
double Closest_Pair(int left,int right)
{
double d = INF;
if(left == right)return d;
if(left + 1 == right)
return dist(p[left],p[right]);
int mid = (left+right)/2;
double d1 = Closest_Pair(left,mid);
double d2 = Closest_Pair(mid+1,right);
d = min(d1,d2);
int k = 0;
for(int i = left;i <= right;i++)
{
if(fabs(p[mid].x - p[i].x) <= d)
tmpt[k++] = p[i];
}
sort(tmpt,tmpt+k,cmpy);
for(int i = 0;i <k;i++)
{
for(int j = i+1;j < k && tmpt[j].y - tmpt[i].y < d;j++)
{
d = min(d,dist(tmpt[i],tmpt[j]));
}
}
return d;
}

int main()
{
//freopen("in.txt","r",stdin) ;
int n;
while(scanf("%d",&n)==1 && n)
{
for(int i = 0;i < n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
sort(p,p+n,cmpxy);
printf("%.2lf\n",Closest_Pair(0,n-1)/2);
}
return 0;
}