Description
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.
看不懂题面没有关系，我来大致讲一下：给定平面上n个黑点和n个白点(就这样认为吧)，求最近的黑点与白点之间的距离。多组输入。输出保留三位小数。
其实这题就是裸的平面最近点对，只需把相同颜色的点之间的距离设为INF即可。
我在这里不做详解，只是稍稍讲一下怎么做。
首先，最近点对 对于一个区间(x坐标)显然 有三种情况：全都在中线左边，全都在中线右边，或者一个在左一个在右。而前两种显然可以递归处理。
那么，我们只需先把点按x坐标排序，然后处理过中线的点对即可。我们定义solve(l,r)来计算在l个点到第r个点之间的最近点对，那么solve(l,r)=min{solve(l,mid),solve(mid+1,r),过中线的最小值}
于是，我们发现在求过中线的最近点对之前我们已经知道了当前的最优解now。由于距离<=now的点离中线的距离一定<=now，那么我们可以把左右两边符合条件的点分别先抠出来。
现在问题变成了：对于任意一个点，是否有点在以它为半径，now为圆心的圆内？
但圆内的点并不好求，我们就可以把圆拓展成矩形：
我们可以发现些什么？是不是最多只有6个点在这些矩形内？
把点再按y轴排一遍序，对于每个点弄个上下边界，一路扫过去即可。总的复杂度为O(nlogn)。
还有一件事：这道题居然卡精度！卡精度！卡精度！请注意你的精度！
当然还有一件非常鬼畜的事(c++选手)：POJ上用printf("%.3lf")时就一直WA，改成printf("%.3f")才可以AC。不知所措。
下面贴代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
#define maxn 200010
#define INF (1LL<<60)
using namespace std;
typedef long long llg;
struct data{
int x,y; bool w;
bool operator < (const data &h)const{return x<h.x;}
}s[maxn],ss[maxn];
llg ans,now;
int n,a[maxn],b[maxn],la,lb,T;
inline llg ji(int x){return (llg)x*(llg)x;}
inline llg dis(int x,int y){ return s[x].w==s[y].w?INF:ji(s[x].x-s[y].x)+ji(s[x].y-s[y].y);}
void work(){
for(int i=1,l=1,r=1;i<=la;i++){
while(s[b[r]].y-s[a[i]].y<now && r<=lb) r++;
while(s[a[i]].y-s[b[l]].y>now && l<=lb) l++;
for(int j=l;j<r;j++) ans=min(ans,dis(a[i],b[j]));
}
}
void solve(int l,int r){
if(l==r) return;
int mid=l+r>>1;
llg m=s[mid].x; now=(llg)sqrt(ans);
solve(l,mid); solve(mid+1,r);
la=0,lb=0;
for(int i=l;i<=mid;i++)
if(m-s[i].x<now) a[++la]=i;
for(int i=mid+1;i<=r;i++)
if(s[i].x-m<now) b[++lb]=i;
work();
int k1=l,k2=mid+1,kk=l-1;
while(k1<=mid && k2<=r)
if(s[k1].y<=s[k2].y) ss[++kk]=s[k1++];
else ss[++kk]=s[k2++];
while(k1<=mid) ss[++kk]=s[k1++];
while(k2<=r) ss[++kk]=s[k2++];
for(int i=l;i<=r;i++) s[i]=ss[i];
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d",&n); ans=INF;
for(int i=1;i<=n*2;i++){
scanf("%d %d",&s[i].x,&s[i].y);
s[i].w= i<=n;
}
sort(s+1,s+n*2+1); solve(1,n<<1);
printf("%.3f
",sqrt(ans));
}
}