POJ-1328-Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54139		Accepted: 12177

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

思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小
岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内斗至少存
在一个点。每次迭代对于第一个区间， 选择最右边一个点， 因为它可以让较多区间得到满足， 如果不选择第一个区间最右一个点（选择前面的点），
那么把它换成最右的点之后， 以前得到满足的区间， 现在仍然得到满足， 所以第一个区间的最右一个点为贪婪选择，选择该点之后，
 将得到满足的区间删掉， 进行下一步迭代， 直到结束。
 不过还要考虑d<0和y>d的情况，island[i].left > temp这个条件比较难搞wa了n次。


#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
using namespace std;
struct node
{
	double left,right;
}island[1001];

bool cmp(node a,node b){
	return a.left < b.left;
}

int main()
{
	double x,y,d,temp;
	int i,cnt,n,k=0;
	bool flag;
	while(scanf("%d%lf",&n,&d)&&!(n==0&&d==0))
	{
		k++;
		flag=false;
		for(i=0;i<n;i++){
			scanf("%lf%lf",&x,&y);
				if(y > d||d<0){
					flag = true;
				}
			island[i].right = x+sqrt(d*d-y*y);
			island[i].left = x-sqrt(d*d-y*y);
		}
		if(flag){
			printf("Case %d: -1\n",k);
			continue;
		}
		sort(island,island+n,cmp);
		temp=island[0].right;
		cnt=1;
		for(i=1;i<n;i++){
			if(island[i].right <= temp){
				temp = island[i].right;
			}
			else if(island[i].left > temp){
				cnt++;
				temp = island[i].right;
			}
		}
		printf("Case %d: %d\n",k,cnt);
	}
	return 0;
}