POJ 1328 - Radar Installation [贪心]

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

#include <cmath>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
using namespace std;

struct Space{
    double l, r;
    Space(double l = 0, double r = 0):l(l),r(r){}
    bool operator < (const Space & rhs) const {
        return r < rhs.r;
    }
};
Space a[1024];

int main()
{
    ios::sync_with_stdio(false);
    int n, kase = 0, ok = 1;
    double d;
    while(memset(a, 0, sizeof(a)), ok = 1, cin >> n >> d && (n || d)){
        for(int i = 0; i < n; ++i){
            double x, y; cin >> x >> y;
            if(y > d){ok = 0; continue;}
            double r = sqrt(d*d - y*y);
            a[i] = Space(x - r, x + r);
        }
        if(!ok){printf("Case %d: %d\n", ++kase, -1); continue;}
        sort(a, a + n);
        double End = a[0].r;
        int cnt = 1;
        for(int i = 1; i < n; ++i){
            if(a[i].l > End ){
                ++cnt;
                End = a[i].r;
            }
        }
        printf("Case %d: %d\n", ++kase, cnt);
    }
    return 0;
}