# POJ 1328 Radar Installation （贪心）

 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 82241 Accepted: 18418

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

### AC 代码

#include <iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#include<queue>
using namespace std;

struct point
{
double x,y;
double s,e;
bool operator<(const point &o)const
{
return e<o.e;
}
} land[1005];

int solve(int n,int d)
{
for(int i=0; i<n; i++)  //计算所有的线段
{
double t=sqrt(d-land[i].y*land[i].y);
land[i].s=land[i].x-t;
land[i].e=land[i].x+t;
}
sort(land,land+n);      //排序
int ans=1;
double r=land[0].e;
for(int i=1; i<n; i++)
{
if(land[i].s>r)
{
++ans;
r=land[i].e;
}
}
return ans;
}

int main()
{
int n,d,cat=1;
while(~scanf("%d%d",&n,&d)&&(n||d))
{
bool flag=false;
for(int i=0; i<n; i++)
{
scanf("%lf%lf",&land[i].x,&land[i].y);
if(land[i].y>d)flag=true;   //是否可以检测到该岛屿
}
if(flag)printf("Case %d: -1\n", cat++);
else printf("Case %d: %d\n", cat++, solve(n,d*d));
}
return 0;
}