解题思路：
分治法求平面近期点对。点分成两部分，加个标记就好了。
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
#include <string.h>
#define LL long long
using namespace std;
const int MAXN = 200000 + 10;
const double INF = 1e100;
struct Point
{
double x, y;
int flag;
}P[MAXN];
int N;
Point vec[MAXN];
bool cmp_x(Point a, Point b)
{
return a.x < b.x;
}
bool cmp_y(Point a, Point b)
{
return a.y < b.y;
}
double dis(Point a, Point b)
{
double dx = a.x - b.x;
double dy = a.y - b.y;
return sqrt(dx * dx + dy * dy);
}
double solve(Point *a, int l, int r)
{
if(l == r) return INF;
if(l + 1 == r)
{
if(P[l].flag == P[r].flag)
return INF;
return dis(P[l], P[r]);
}
int m = (l + r) >> 1;
double d = solve(a, l, m);
d = min(d, solve(a, m + 1, r));
int sz = 0;
for(int i=l;i<=r;i++)
{
if(fabs(P[i].x - P[m].x) <= d)
vec[sz++] = P[i];
}
sort(vec, vec + sz, cmp_y);
for(int i=0;i<sz;i++)
{
for(int j=i+1;j<sz;j++)
{
if(fabs(vec[i].y - vec[j].y) >= d)
break;
if(vec[i].flag != vec[j].flag)
{
double rs = dis(vec[i], vec[j]);
if(rs < d) d = rs;
}
}
}
return d;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d", &N);
for(int i=0;i<N;i++)
{
scanf("%lf%lf", &P[i].x, &P[i].y);
P[i].flag = 0;
}
for(int i=0;i<N;i++)
{
scanf("%lf%lf", &P[i + N].x, &P[i + N].y);
P[i + N]. flag = 1;
}
N <<= 1;
sort(P, P + N, cmp_x);
double ans = solve(P, 0, N - 1);
printf("%.3f
", ans);
}
return 0;
}