HDOJ-三部曲-1002-Radar Installation

Radar Installation

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 60   Accepted Submission(s) : 11

Problem 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

3 2

1 2

-3 1

2 1

 

1 2

0 2

 

0 0

 

Sample Output

Case 1: 2
Case 2: 1

 

Source

PKU

 

 

 

今天通过这道题又学习了一下贪心算法的知识，感觉收获很多。

对于每个岛屿的位置，如果y大于d则雷达无法侦测到岛屿，结果为-1；如果y都小于d，将每个岛屿按照x坐标从小到大排列，并算出对于每个岛屿，一个雷达能侦测到这个岛屿时雷达在x轴上的最左和最右的位置，并储存为lx和rx。从最左边的岛屿算起，如果该岛屿（记为a1）右边下一个岛屿（记a2）的lx大于a1的rx，则只能再加一个雷达才能侦测到a2；如果a2的lx小于a1的rx，且a2的rx小于a1的rx则a2的再下一个岛屿（记为a3）的lx必须小于a2的rx才能被原来的雷达检测到，否则要再加一个雷达（即a3必须和a2的rx比较）；如果a2的lx小于a1的rx，且a2的rx大于a1的rx，能侦测到两个岛屿的雷达坐标范围是a2的lx到a1的rx，则a3必须和a1的rx进行比较来判断是否应该再加一个雷达。

 

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
struct island
{
	int x,y;
	double rx,lx;
};

int cmp(const island &a,const island &b)
{
	if(a.x<b.x)
		return 1;
	else
		return 0;
}

int main()
{
	int cas=0,n,d;
	while(cin>>n>>d&&n+d)
	{
		cas++;
		island is[1001];
		bool f=true;
		for(int i=0;i<n;i++)
		{
			cin>>is[i].x>>is[i].y;
			if(is[i].y>d)
				f=false;
			double t=sqrt(d*d-is[i].y*is[i].y);
			is[i].lx=is[i].x-t;
			is[i].rx=is[i].x+t;
		}
		if(!f)
		{
			cout<<"Case "<<cas<<": "<<-1<<endl;
		}
		else
		{
			sort(is,is+n,cmp);
			/*for(int i=0;i<n;i++)
				cout<<is[i].x<<' '<<is[i].rx<<' '<<is[i].lx<<endl;*/
			double temp=is[0].rx;
			int count=1;
			for(int i=1;i<n;i++)
			{
				if(is[i].lx>temp)
				{
					count++;
					temp=is[i].rx;
				}
				else if(is[i].rx<temp)
					temp=is[i].rx;
			}
			cout<<"Case "<<cas<<": "<<count<<endl;
		}
	}
}