1. 程式人生 > >【分治法】最接近點對問題——Java 實現

【分治法】最接近點對問題——Java 實現

問題描述

       給定平面上n個點,找其中的一對點,使得在n個點組成的所有點對中,該點對間的距離最小。

       注:
a、嚴格地講,最接近點對可能多餘1對,為簡單起見,只找其中的1對作為問題的解。
b、一個簡單的演算法是——只要將每一個點與其他 n-1 個點的距離算出,找到達到最小距離的2個點即可。然而,這樣做效率太低,需要O(n^2)的計算時間 c、已經證明,該問題的計算時間下界為Ω(nlogn)。

演算法測試實現:


Jimport javax.swing.JOptionPane;

public class ClosestPair {
	//合併排序演算法
	//定義要排序的的陣列
	public static class MergeSort{
		//排序演算法
		public static void mergeSort(Comparable a[]){
			Comparable b[] = new Comparable[a.length];
			int s = 1;
			while(s < a.length){
				mergePass(a,b,s);//合併到陣列b
				s += s;
				mergePass(b,a,s);//合併到陣列a
				s += s;
			}
		}
		//合併大小為s的相鄰子陣列
		public static void mergePass(Comparable x[],Comparable y[],int s){
			int i = 0;
			while(i <= x.length - 2 * s){
				//合併大小為s的相鄰2段子陣列
				merge(x,y,i,i + s - 1,i + 2 * s - 1);
				i = i + 2 * s;
			}
			if(i + s < x.length)
				merge(x,y,i,i + s - 1,x.length - 1);
			else
				//複製到y
				for(int j = i;j < x.length;j ++)
					y[j] = x[j];
		}
		public static void merge(Comparable c[],Comparable d[],int l,int m,int r){
			//合併 c[l:m]和c[m+1:r]到d[l:r]
			int i = l,j = m + 1,k = l;
			while((i <= m) && (j <= r))
				if(c[i].compareTo(c[j]) <= 0){
					d[k ++] = c[i ++];
				}
				else
					d[k ++] = c[j ++];
			if(i > m)
				for(int q = j;q <= r;q ++)
					d[k ++] = c[q];
			else
				for(int q = i;q <= m;q ++)
					d[k ++] = c[q];
		}
	}
	public static class Point{
		double x,y;
		public Point(double xx,double yy){
			x = xx;
			y = yy;
		}
	}
	public static class Point1 extends Point implements Comparable{
		int id;
		public Point1(double xx,double yy,int theID){
			super(xx,yy);
			id = theID;
		}
		public int compareTo(Object x){
			double xx = ((Point1)x).x;
			if(this.x < xx)
				return -1;
			if(this.x == xx)
				return 0;
			return 1;
		}
		public boolean equals(Object x){
			return this.x == ((Point1)x).x;
		}
		public void getID(){
			System.out.println(id);
		}
	}
	public static class Point2 extends Point implements Comparable{
		int p;
		public Point2(double xx,double yy,int pp){
			super(xx,yy);
			p = pp;
		}
		public int compareTo(Object x){
			double xy = ((Point2)x).y;
			if(this.y < xy)
				return -1;
			if(this.y == xy)
				return 0;
			return 1;
		}
		public boolean equals(Object x){
			return this.y == ((Point2)x).y;
		}
		//輸出點的序號
		public void getp(){
			System.out.println(p);
		}
	}
	public static class Pair{
		Point1 a;
		Point1 b;
		double dist;
		public Pair(Point1 aa,Point1 bb,double dd){
			a = aa;
			b = bb;
			dist = dd;
		}
		public void print(){
			System.out.println("輸出點a的序號:");
			a.getID();
			System.out.println("輸出點b的序號:");
			b.getID();
			System.out.println("輸出兩點之間的距離:");
			System.out.println(dist);
		}
	}
	public static double dist(Point u,Point v){
		double dx = u.x - v.x;
		double dy = u.y - v.y;
		return Math.sqrt(dx * dx + dy * dy);
	}
	public static Pair cpair2(Point1 x[]){
		if(x.length < 2)
			return null;
		MergeSort.mergeSort(x);
		Point2 y[] = new Point2[x.length];
		for(int i=0;i < x.length;i ++)
			//將陣列中x中的點複製到陣列y中
			y[i] = new Point2(x[i].x,x[i].y,i);
		MergeSort.mergeSort(y);//依y排序
		Point2 z[] = new Point2[x.length];
		//計算最近點對
		return closestPair(x,y,z,0,x.length - 1);
	}
	
	private static Pair closestPair(Point1 x[],Point2 y[],Point2 z[],int l,int r){
		if(r - l == 1)//2點的情形
			return new Pair(x[l],x[r],dist(x[l],x[r]));
		if(r - l == 2){
			//三點的情形
			double d1 = dist(x[l],x[l + 1]);
			double d2 = dist(x[l + 1],x[r]);
			double d3 = dist(x[l],x[r]);
			if(d1 <= d2 && d1 <= d3)
				return new Pair(x[l],x[l + 1],d1);
			if(d2 <= d3)
				return new Pair(x[l + 1],x[r],d2);
			else
				return new Pair(x[l],x[r],d3);
		}
		//多於3點的情形,用分治法
		int m = (l + r) / 2;
		int f = l,g = m + 1;
		for(int i = l;i <= r;i++)
			if(y[i].p > m)
				z[g ++] = y[i];
			else z[f ++] = y[i];
		//遞迴求解
		Pair best = closestPair(x,z,y,l,m);
		Pair right = closestPair(x,z,y,m + 1,r);
		if(right.dist < best.dist)
			best = right;
		MergeSort.merge(z,y,l,m,r);//重構陣列y
		//d矩形條內的點置於z中
		int k = l;
		for(int i = l;i <= r;i ++)
			if(Math.abs(x[m].x - y[i].x) < best.dist)
				z[k ++] = y[i];
		//搜尋z[l:k-1]
		for(int i = l;i < k;i++){
			for(int j = i + 1;j < k && z[j].y - z[i].y < best.dist;j ++){
				double dp = dist(z[i],z[j]);
				if(dp < best.dist)
					best = new Pair(x[z[i].p],x[z[j].p],dp);
			}
		}
		return best;
	}
	public static void main(String[]args){
		String m = JOptionPane.showInputDialog(null,"輸入點的個數:","最接近點問題",JOptionPane.QUESTION_MESSAGE);
		int n = Integer.parseInt(m);
		Point1 p[] = new Point1[n];
		for(int i = 0;i < n;i ++){
			String s1=JOptionPane.showInputDialog(null,"輸入p["+i+"]的橫座標","最接近點問題",JOptionPane.QUESTION_MESSAGE);
			int x = Integer.parseInt(s1);
			String s2 = JOptionPane.showInputDialog(null,"輸入p["+i+"]的縱座標",
					"最接近點問題",JOptionPane.QUESTION_MESSAGE);
			int y = Integer.parseInt(s2);
			p[i] = new  Point1(x,y,i);
		}
		Pair pair = cpair2(p);
		pair.print();
	}
}