POJ 1328 Radar Installation

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本文探讨了在海岸线上安装雷达以覆盖海中岛屿的问题。通过贪心算法确定最少雷达数量，实现全面覆盖。考虑到坐标排序和边界条件，确保了算法的有效性和准确性。

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36401		Accepted: 8092

Description

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.

The input is terminated by a line containing pair of zeros

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

Sample Output

Case 1: 2
Case 2: 1

Source

Beijing 2002

解题思路：貌似很有名的一题，之前听人讲过，用的是贪心，对于每个岛屿其雷达点可设在[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)]的区间内，先对所有区间的左坐标进行排序（这里我欺负POJ测试数据弱就没有用快排了），然后设置雷达位置为第一个区间的右坐标，若雷达位置在后一个区间的左坐标之前（即两区间没有交集），就增设一个雷达，同时把该雷达部署到后一个区间的右坐标位置，否则，若后一个区间右坐标小于原雷达位置，就把原雷达改为部署到后一个区间的右坐标位置，不然，则原雷达位置不变。需要注意的两点是边界条件的判断和精度的确保，对前者大家说法不一，而我则是把y<0,d<=0,d<y,三种情况都加以考虑，后者关键是用于存区间的数组，点的坐标，雷达位置均要用浮点型表示（我在这里WA一次）。

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int i,j,island,ans,test=1;
	double x,y,d,R,*l,*r;
	bool flag;
	while(cin>>island>>d&&(island||d))
	{
		ans=1;flag=true;
		if(d<=0) flag=false;
		l=new double[island+1];
		r=new double[island+1];
		memset(l,0,sizeof(l));
		memset(r,0,sizeof(r));
		for(i=0;i<island;i++)
		{
			cin>>x>>y;
			if(y>d||y<0) flag=false;
			l[i]=x-sqrt(1.0*d*d-y*y);
			r[i]=x+sqrt(1.0*d*d-y*y);
		}
		if(flag)
		{
			for(i=0;i<island;i++)
			for(j=i+1;j<island;j++)
				if(l[i]>l[j])
				{
					swap(l[i],l[j]);
					swap(r[i],r[j]);
				}
				R=r[0];
		for(i=1;i<island;i++)
		{
			if(R<l[i])
				{
					ans++;
					R=r[i];
			    }
			else if(R>r[i])
			    R=r[i];
		}
		    cout<<"Case "<<test<<": "<<ans<<endl;
		}
		else
			cout<<"Case "<<test<<": "<<-1<<endl;
		test++;
	}
	delete l;
	delete r;
	return 0;
}