POJ - 1328 Radar Installation

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

以各个小岛为圆心，d为半径做圆，与x轴的两个交点记录，排序，贪心从左到右找出最少需要的雷达。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>

using namespace std;

struct node
{
    double l,r;
}c[1005];

bool cmp(node a,node b)
{
    return a.l<b.l;
}

int main()
{
    int n,d,x,y,s=1;
    while(scanf("%d%d",&n,&d),n||d)
    {
        int flag=0;
        memset(c,0,sizeof(c));
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            if(flag) continue;
            if(y>d)
                flag=1;
                else
                {
                    c[i].l=(double)x-sqrt((double)d*d-y*y);
                    c[i].r=(double)x+sqrt((double)d*d-y*y);
                }
        }
        if(flag)
        {
            printf("Case %d: -1\n",s++);
            continue;
        }
        sort(c,c+n,cmp);
        int ans=1;
        double h=c[0].r;
        for(int i=1;i<n;i++)
        {
            if(c[i].r<h)
                h=c[i].r;
            else if(c[i].l>h)
            {
                h=c[i].r;
                ans++;
            }
        }
        printf("Case %d: %d\n",s++,ans);
    }
    
    
    return 0;
}