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题目链接：http://poj.org/problem?id=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
思路：反向思考把每个岛屿能到达的左区间和右区间用结构体记录，再用贪心的思想计算雷达数；
代码如下：
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
struct point
{
double r,l;
}P[1017];
bool cmp(point a,point b)
{
return a.l < b.l;
}
int main()
{
int n, m, r, flag;
int i, j;
int x, y;
int cas = 0, cont;
while(cin>>n>>r)
{
if(n == 0 && r == 0)
break;
flag = cont = 0;
for(i = 0; i < n; i++)
{
cin>>x>>y;
if(fabs(y) > r)
{
flag = 1;
}
P[i].l = x*1.000-sqrt(r*r*1.00-y*y*1.000);
P[i].r = x*1.000+sqrt(r*r*1.00-y*y*1.000);
}
cout<<"Case "<<++cas<<": ";
if(flag)
{
cout<<"-1"<<endl;
continue;
}
sort(P,P+n,cmp);
cont = 1;
double t = P[0].r;
for(i = 1; i < n; i++)
{
if(P[i].r < t)//为了防止上一个岛屿的右区间比下一个岛屿的右区间还大的情况
{//如果岛屿的右区间在雷达的左边，就需要移动雷达的位置到此岛屿的右区间
t = P[i].r;
}
if(P[i].l > t)//如果下一个海岛的左区间在雷达的右边
{//就需要再安一个雷达在此岛屿的有区间
t = P[i].r;
cont++;
}
}
cout<<cont<<endl;
}
return 0;
}
poj1328 Radar Installation（贪心）