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HDU1007 Quoit Design 最近点对

HDU1007 Quoit Design http://acm.hdu.edu.cn/showproblem.php?pid=1007 字典树问题

套用吉林大学的模板水过。

题目大意:给定一个平面的很多点(最多10万个),要求求出其中相邻距离最近的两个点之间的距离的一半。暴力计算肯定超时,需要分治解决。

//==============================================
// Name        : HDU1007
// Author      : 代码疯子
// Version     : 最近点对
// Copyright   : http://www.programlife.net
// Description : Quoit Design
//==============================================

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
using namespace std;

//////////////////////////////////////////////
const int N = 100005;
const double MAX = 10e100, eps = 0.00001;

typedef struct TagPoint
{
	double x;
	double y;
	int index;
}Point;

Point a[N], b[N], c[N];

double closest(Point *, Point *, Point *, int, int);
double dis(Point, Point);
int cmp_x(const void *, const void*);
int cmp_y(const void *, const void*);
int merge(Point *, Point *, int, int, int);
inline double min(double, double);

double closest(Point a[],Point b[],Point c[],int p,int q)
{
	if (q - p == 1) return dis(a[p], a[q]);
	if (q - p == 2) 
	{
		double x1 = dis(a[p], a[q]);
		double x2 = dis(a[p + 1], a[q]);
		double x3 = dis(a[p], a[p + 1]);
		if (x1 < x2 && x1 < x3) return x1;
		else if (x2 < x3) return x2;
		else return x3;
	}
	int i, j, k, m = (p + q) / 2;
	double d1, d2;
	for (i = p, j = p, k = m + 1; i <= q; i++)
		if (b[i].index <= m) c[j++] = b[i];
	//数组c左半部保存划分后左部的点, 且对y是有序的.
		else c[k++] = b[i];
		d1 = closest(a, c, b, p, m);
		d2 = closest(a, c, b, m + 1, q);
		double dm = min(d1, d2);
		//数组c左右部分分别是对y坐标有序的, 将其合并到b.
		merge(b, c, p, m, q);
		for (i = p, k = p; i <= q; i++)
			if (fabs(b[i].x - b[m].x) < dm) c[k++] = b[i];
		//找出离划分基准左右不超过dm的部分, 且仍然对y坐标有序.
		for (i = p; i < k; i++)
			for (j = i + 1; j < k && c[j].y - c[i].y < dm; j++)
			{
				double temp = dis(c[i], c[j]);
				if (temp < dm) dm = temp;
			}
			return dm;
}
double dis(Point p, Point q)
{
	double x1 = p.x - q.x, y1 = p.y - q.y;
	return sqrt(x1 *x1 + y1 * y1);
}

int merge(Point p[], Point q[], int s, int m, int t)
{
	int i, j, k;
	for (i=s, j=m+1, k = s; i <= m && j <= t;) 
	{
		if (q[i].y > q[j].y) 
			p[k++] = q[j], j++;
		else
			p[k++] = q[i], i++;
	}
	while (i <= m) p[k++] = q[i++];
	while (j <= t) p[k++] = q[j++];
	memcpy(q + s, p + s, (t - s + 1) *sizeof(p[0]));
	return 0;
}

int cmp_x(const void *p, const void *q)
{
	double temp = ((Point*)p)->x - ((Point*)q)->x;
	if (temp > 0)
		return 1;
	else if (fabs(temp) < eps)
		return 0;
	else 
		return - 1;
}

int cmp_y(const void *p, const void *q)
{
	double temp = ((Point*)p)->y - ((Point*)q)->y;
	if(temp > 0)
		return 1;
	else if (fabs(temp) < eps)
		return 0;
	else
		return - 1;
}

inline double min(double p, double q)
{
	return (p > q) ? (q): (p);
}

int main()
{
	int n, i;
	double d;
	while(EOF != scanf("%d", &n), n)
	{
		for (i = 0; i < n; ++i)
		{
			scanf("%lf %lf", &a[i].x, &a[i].y);
		}
		qsort(a, n, sizeof(a[0]), cmp_x);
		for (i = 0; i < n; i++)
		{
			a[i].index = i;
		}
		memcpy(b, a, n * sizeof(a[0]));
		qsort(b, n, sizeof(b[0]), cmp_y);
		d = closest(a, b, c, 0, n - 1);
		printf("%.2f\n", d / 2.0);
	}
	return 0;
}

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