poj 1328 【Radar Installation】【几何转化、区间覆盖】

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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

Output

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

求：在x坐标轴上，至少要画几个以d为半径的圆，才能使P点均在圆内。

解题思路：分别以P点为圆心，d为半径画圆，圆分别会在x轴上截取相应的线段，只要线段内有圆心，则即可覆盖P点，因此问题

便转化为求X轴上共有多少独立区间！

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
using namespace std;
struct A{
	double zuo,you;
};
int cmp(A a,A b){
	return a.zuo<b.zuo;
}
int main(){
	int t=1,n,i,r,x,y,yes;
	A a[1200];
	while(scanf("%d%d",&n,&r)&&!(n==0&&r==0)){
        yes=0;
		for(i=1;i<=n;i++){
			scanf("%d%d",&x,&y);
			if(r<y) yes=1;
			double num=sqrt((double)(r*r-y*y));
			a[i].zuo=double(x-num);
			a[i].you=double(x+num);
		}
		if(yes){printf("Case %d: -1\n",t++);continue;}
		sort(a+1,a+n+1,cmp);
		double k=a[1].you;
		int count=1;
		for(i=2;i<=n;i++){
			if(a[i].zuo>k){
				k=a[i].you;
				count++;
			}
			else{
				if(a[i].you<k)
					k=a[i].you;
			}
		}
		printf("Case %d: %d\n",t++,count);
	}
	return 0;
}