POJ平面最近点对

描述

二维平面上有N个点，求最近点对之间的距离。

输入第一行一个整数T，表示有T组测试数据
每组测试数据第一行一个整数N(2<=N<=1e5)表示平面有N个点
接下来有N行，每行两个整数X Y(-1e9<=X,Y <=1e9)表示点的坐标输出输出最近点对的距离，精确到小数点后6位样例输入

1
3
1 0
1 1
0 1

样例输出

1.000000

#include <iostream>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <iostream>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>

#define INF 0x3f3f3f3f

using namespace std;

struct point{
    double x, y;
    point(double x, double y) :x(x), y(y){};
};
vector<point> vr;
bool cmpx(point a, point b){
    return a.x < b.x;
}
bool cmpy(int a, int b){
    return vr[a].y < vr[b].y;
}
double getdis(point a,point b){
    return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));
}
double mind(double a, double b){
    return a < b ? a : b;
}

double fun(int left, int right){
    if (left + 1 == right) return getdis(vr[left], vr[right]);
    if (left + 2 == right)return mind(getdis(vr[left], vr[left + 1]), mind(getdis(vr[left], vr[right]), getdis(vr[left + 1], vr[right])));
    int mid = (left + right) >> 1;
    double d = mind(fun(left, mid), fun(mid + 1, right));
    vector<int>tv;
    for (int i = left; i <= right; i++){
        if (abs(vr[i].x - vr[mid].x) < d) tv.push_back(i);
    }
    sort(tv.begin(), tv.end(), cmpy);
    for (int i = 0; i < tv.size()-1; i++)
    for(int j=i+1;j<tv.size();j++){
        if (abs(vr[tv[i]].y - vr[tv[j]].y) < d)
            d = mind(d, getdis(vr[tv[i]], vr[tv[j]]));
    }
    return d;
}


int main()
{
    int T,N;
    double a, b;
    scanf("%d", &T);
    while (T--){
        vr.clear();
        scanf("%d", &N);
        for (int i = 0; i < N; i++){
            scanf("%lf %lf", &a, &b);
            vr.push_back(point(a, b));
        }
        sort(vr.begin(), vr.end(), cmpx);
        printf("%.6lf\n",fun(0, N - 1));
    }
    return 0;
}