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
题意：雷达只能建造在海岸线上（水平的X轴），且所有雷达的覆盖半径是相同的。现在问是否可以建造最少的雷达使得所有的小岛都被覆盖。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include<math.h>
using namespace std;
const int maxn = 1e5+10;
typedef long long ll;
int a[maxn];
struct PIONT {
    double st, ed;
}p[maxn];
bool cmp(const PIONT &a, const PIONT &b)
{
    if (a.st == b.st)a.ed < b.ed;
    return a.st< b.st;
}
int main()
{
    int n, d;
    int cas = 0;
    while(~scanf("%d%d",&n,&d)&&n&&d){
        int flag = 0;
        for (int i = 1; i <= n; ++i)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            if (y > d)flag = 1;
            else
            {
                double s = sqrt(d * d - y * y);
                p[i].st = x - s;
                p[i].ed = x + s;
            }
        }
            if (flag)printf("Case %d: -1\n", ++cas);
            else
            {
                sort(p + 1, p + 1 + n, cmp);
                double r = p[1].ed;
                int ans = 1;
                for (int i = 2; i <= n; ++i)
                {
                    if (p[i].st <= r) r = min(r, p[i].ed);
                    else 
                    { 
                        ans++;
                        r = p[i].ed;
                    }
                }
                printf("Case %d: %d\n", ++cas, ans);
            }
        }
    return 0;
}