POJ - 3714 Raid （平面分治最小点对）

After successive failures in the battles against the Union, the Empire retreated to its last stronghold. Depending on its powerful defense system, the Empire repelled the six waves of Union's attack. After several sleepless nights of thinking, Arthur, General of the Union, noticed that the only weakness of the defense system was its energy supply. The system was charged by N nuclear power stations and breaking down any of them would disable the system.

The general soon started a raid to the stations by N special agents who were paradroped into the stronghold. Unfortunately they failed to land at the expected positions due to the attack by the Empire Air Force. As an experienced general, Arthur soon realized that he needed to rearrange the plan. The first thing he wants to know now is that which agent is the nearest to any power station. Could you, the chief officer, help the general to calculate the minimum distance between an agent and a station?

 

Input

The first line is a integer T representing the number of test cases.
Each test case begins with an integer N (1 ≤ N ≤ 100000).
The next N lines describe the positions of the stations. Each line consists of two integers X (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the station.
The next following N lines describe the positions of the agents. Each line consists of two integers X (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the agent. 　

 

Output

For each test case output the minimum distance with precision of three decimal placed in a separate line.

 

Sample Input

2
4
0 0
0 1
1 0
1 1
2 2
2 3
3 2
3 3
4
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0

 

Sample Output

1.414
0.000

题目大意：给出n个核电站的位置和n个特工的位置，输出其中特工和核电站最短的距离

解题思路：平面分治最小点对的模板题，不同的是需要将核电站和特工的坐标标记分开，

平面分治最小点对： https://blog.youkuaiyun.com/qq_40707370/article/details/85268256

AC代码：

#include<iostream> 
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=1e6;
const double INF=1e100; 
struct node{
	double x,y;
	int flag;
}a[maxn],term[maxn];
bool cmpx(node a,node b)
{
	return a.x<b.x;
}
bool cmpy(node a,node b)
{
	return a.y<b.y;
}
double dist(node a,node b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double mindist(int left,int right)
{
	if(left==right)
		return INF;
	if(left+1==right)
	{
		if(a[left].flag==a[right].flag)
			return INF;
		else
			return dist(a[left],a[right]);
	}
	int mid=(left+right)/2;
	double d1=mindist(left,mid);//左区间最小值 
	double d2=mindist(mid+1,right);//右区间最小值 
	double dis=min(d1,d2);
	int i,j,k=0;
	for(i=left;i<=right;i++)
	{
		if(fabs(a[i].x-a[mid].x)<=dis)
			term[k++]=a[i];
	}
	sort(term,term+k,cmpy);
	for(i=0;i<k;i++)
	{
		for(j=i+1;j<k;j++)
		{
			if(term[j].y-term[i].y>dis)
				break;
			if(term[i].flag!=term[j].flag)
				dis=min(dis,dist(term[i],term[j]));
		}
	} 
	return dis;
}
int main()
{
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%lf%lf",&a[i].x,&a[i].y);
			a[i].flag=0;
		}	
		for(int i=0;i<n;i++)
		{
			scanf("%lf%lf",&a[n+i].x,&a[n+i].y);
			a[n+i].flag=1;
		}
		n*=2; 
		sort(a,a+n,cmpx);
		double ans=mindist(0,n-1);
		printf("%.3lf\n",ans);
	}
	return 0;
}