Radar Installation(贪心)

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
题意：有一个坐标轴，在x轴的上方时海，下方是陆地，x轴为海岸线，海上有n个小岛，因为种种原因要在海岸线上装雷达，每个雷达的覆盖范围是以r为半径的圆，请用最少的雷达数覆盖所有的小岛，当无法完全覆盖时输出-1；
思路：典型的贪心思想。以小岛为圆心以r为半径做园，计算出圆与x轴的左交点和右交点。左交点为x-sqrt（r*r-y*y），右交点为x+sqrt（r*r+y*y）；然后对左交点从小到大排序，令初始雷达为最小岛屿的右交点，如果i点的左交点在雷达的右面，则需要重新装一个雷达，然后令当前点为新雷达的右交点，否则如果i点的右交点在当前雷达的左边，则把当前雷达的位置更新为该点的右交点。
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
struct node
{
    double l,r;
}point[1010];
int cmp(struct node a,struct node b)//对左交点进行排序；
{
    return a.l<b.l;
}
int main()
{
    int n,r,i;
    double x,y,t;//用t来储存当前雷达的位置
    int cnt=1;
    while(~scanf("%d %d",&n,&r))
    {
        if(n==0&&r==0)
            break;
        int ans=1;//记录安装雷达的个数
        for(i=0;i<n;i++)
        {
            scanf("%lf %lf",&x,&y);
            point[i].l=x-sqrt(r*r-y*y);//当前点与x轴的左交点
            point[i].r=x+sqrt(r*r-y*y);//当前点与x轴的右交点
            if(y>r||y<0||r<=0)//列举出三种无法覆盖的情况
                ans=-1;
        }
        sort(point,point+n,cmp);//排序
        t=point[0].r;//令初始值为位置最小岛屿的右交点
        for(i=1;i<n&&ans!=-1;i++)
        {
            if(point[i].l>t)
            {
                ans++;
                t=point[i].r;
            }
           if(point[i].r<t)
                {
                    t=point[i].r;
                }
        }
        printf("Case %d: %d\n",cnt++,ans);
    }
    return 0;
}