Radar Installation（POJ-1328）

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

题意：海上有很多岛屿，每个岛屿都给你坐标，让你在x轴上（相当于海岸线）放置雷达，最少放几个；

思路：贪心，把每个岛屿放置雷达的区间求出来，沿途比较就行；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define INF 0x3f3f3f3f
#define maxn 1010
using namespace std;
struct edge
{
double l,r;
};
edge g[maxn];
bool cmp(edge a,edge b)
{
return a.l<b.l;
}
double x,y;
int n,d,t;
int main()
{
t=0;
while(scanf("%d%d",&n,&d)!=EOF){
if(n==0&&d==0) break;
t++;
bool flag=true;
for(int i=0;i<n;i++){
scanf("%lf%lf",&x,&y);
if(y>d) flag=false;
g[i].l=x-sqrt(1.0*d*d-y*y);
g[i].r=x+sqrt(1.0*d*d-y*y);
}
printf("Case %d: ",t);
if(!flag){
cout<<-1<<endl;
continue;
}
sort(g,g+n,cmp);
double xx=g[0].r;
int sum=1;
for(int i=1;i<n;i++){
if(g[i].r<xx) xx=g[i].r;
else if(g[i].l>xx){
sum++;
xx=g[i].r;
}
}
cout<<sum<<endl;
}
}