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 InstallationsInput
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
以每个点为圆心画圆，如果和x轴没交点表示没有方案，答案为-1，否则记录与x轴的两个交点（可能是一个，按两个来处理），然后按右交点升序排序，
排着遍历，取出一个圆，然后后面的圆左交点小于等于这个圆的右交点的，都可以用一个雷达感应到。
说一下排序的原因为啥是以右交点升序，而不是降序，也不是以左点，因为是从左往右看，一个点越靠近x轴，那么两交点的距离越大，那么可能他左边有一些点可以共用一个雷达，右边有一些点也可以共用一个雷达，但是这些点都在这个点所成圆的两交点之间，却不能共用一个雷达。
如图ABC三点，如果按左点排序不管右点，那么如果A点排在前面，BC的左点都不超过A的右点，实际需要两个雷达，显然B要排在前面。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define MAX 1001
#define inf 0x3f3f3f3f
using namespace std;
typedef pair<double,double> pa;
int n,num,c;
double r;
pa p[MAX];
bool cmp(pa a,pa b) {
return a.second < b.second;
}
int main() {
double a,b;
while(~scanf("%d%lf",&n,&r) && (n + r)) {
int flag = 1;
for(int i = 0;i < n;i ++) {
scanf("%lf%lf",&a,&b);
if(fabs(b) > r)flag = 0;
else {
double d = sqrt(r * r - b * b);
p[i].first = a - d;
p[i].second = a + d;
}
}
if(!flag)c = -1;
else c = 0;
if(!c) {
sort(p,p + n,cmp);
int i = 0;
while(i < n) {
int j = i + 1;
while(j < n && p[j].first <= p[i].second)j ++;
c ++;
i = j;
}
}
printf("Case %d: %d\n",++ num,c);
}
}
poj 1328 Radar Installation
原文：https://www.cnblogs.com/8023spz/p/9520912.html
