【贪心】【POJ-1328&AOJ-195】Radar Installation

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

/***********************************************************************************************************************
题意：有n个岛屿和雷达覆盖半径d 求出要覆盖这些岛屿最少需要多少雷达
思路：把每个岛屿想象成一个雷达中心 则在岸边的覆盖区间可以写出 根据区间是否重叠来求出需要多少雷达
***********************************************************************************************************************/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
typedef struct 
{
    double l, r;
}node;
node a[1000+10];
bool cmp(node a , node b)
{
    return a.l < b.l;
}
int main()
{
    //freopen("data.in" , "r" , stdin);
    int n , d , cas = 0;
    while(scanf("%d %d", &n , &d) && (n || d))
    {
        bool run = true;
        for(int i = 0 ; i < n ; i ++)
        {
            int tx, ty;
            scanf("%d %d", &tx, &ty);
            if(abs(ty) > d)
                run = false;
            a[i].l = (double)tx - sqrt((double)(d * d - ty * ty));
            a[i].r = (double)tx + sqrt((double)(d * d - ty * ty));
        }
        if(!run)
        {
            printf("Case %d: -1\n" , ++cas);
            continue;
        }
        sort(a , a + n , cmp);
        int ans = 1;
        double last = a[0].r;
        for(int i = 1 ; i < n ; i ++)
        {
            if(a[i].l > last)
            {
                ans ++;
                last = a[i].r;
            }
            else if (a[i].r < last)
                last = a[i].r;
        }
        printf("Case %d: %d\n" , ++cas , ans);
    }
    return 0;
}