HD-ACM算法专攻系列（15）——Quoit Design

Posted 2020-10-09 forcheng

问题描述：

 

 

源码：

经典问题——最近邻问题，标准解法

#include"iostream"
#include"algorithm"
#include"cmath"
using namespace std;

struct Point
{
	double x;
	double y;
};

Point S[100000];//不使用全局变量可能会超内存

bool cmpPointX(Point a, Point b)
{
	return a.x > b.x;
}

bool cmpPointY(Point a, Point b)
{
	return a.y > b.y;
}

double EfficientClosestPair(Point *P, Point *Q, int start, int n)
{
	if(n <= 3)
	{
		double result;
		result = (P[start].x - P[start+1].x)*(P[start].x - P[start+1].x) + (P[start].y - P[start+1].y) * (P[start].y - P[start+1].y);
		if(n == 3)
		{
			double tmp = (P[start].x - P[start+2].x)*(P[start].x - P[start+2].x) + (P[start].y - P[start+2].y) * (P[start].y - P[start+2].y);
			if(tmp < result)result = tmp;
			tmp = (P[start+1].x - P[start+2].x)*(P[start+1].x - P[start+2].x) + (P[start+1].y - P[start+2].y) * (P[start+1].y - P[start+2].y);
			if(tmp < result)result = tmp;
		}
		return result;
	}
	else
	{
		int half = n / 2;
		double d1 = EfficientClosestPair(P, Q, start, half);
		double d2 = EfficientClosestPair(P, Q, start + half, n - half);
		double d = d1 < d2 ? d1 : d2;
		double m = P[start + half - 1].x;
		int count = 0;
		double tmp;
		for(int i = start; i < start + n; i++)
		{
			tmp = Q[i].x - m;
			if(tmp < 0)tmp = - tmp;
			if(tmp < d)count++;
		}
		double dminsq = d;
		if(count > 1)
		{
			//Point *S = new Point[count];
			for(int i = start, j = 0; i < start + n; i++)
			{
				tmp = Q[i].x - m;
				if(tmp < 0)tmp = - tmp;
				if(tmp < d)
				{
					S[j].x = Q[i].x;
					S[j].y = Q[i].y;
					j++;
				}
			}
			int k;
			for(int i = 0; i < count - 1; i++)
			{
				k = i + 1;
				while(k < count && (S[k].y - S[i].y)*(S[k].y - S[i].y) < dminsq)
				{
					dminsq = min((S[k].x - S[i].x)*(S[k].x - S[i].x) + (S[k].y - S[i].y)*(S[k].y - S[i].y), dminsq);
					k++;
				}
			}
		}
		
		return dminsq;
		
	}
}

int main()
{
	int n;
	Point *P, *Q;
	cout.precision(2);
	cout.setf(ios::fixed);
	P = new Point[100000];
	Q = new Point[100000];
	while(scanf("%d", &n) != EOF)
	{
		if(n == 0)break;
		for(int i = 0; i < n; i++)
		{
			scanf("%lf %lf", &P[i].x,&P[i].y);
			Q[i].x = P[i].x;
			Q[i].y = P[i].y;
		}
		if(n <= 3)
		{
			cout<<sqrt(EfficientClosestPair(P, Q, 0, n)) / 2<<endl;
		}
		else
		{
			sort(P, P+n, cmpPointX);//使用qsort可能会超时
			sort(Q, Q+n, cmpPointY);
			cout<<sqrt(EfficientClosestPair(P, Q, 0, n)) / 2<<endl;
		}
	}
    return 0;
}