Radar Installation

B - 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：给出岛屿的坐标和雷达的辐射范围，在x轴上求出能够将所有岛屿覆盖所需的最小雷达数，若不能完成则输出-1。

题解：求出各个岛屿到x轴的距离，再由此求出在x轴上可行的范围，再将范围排序，从左到右安放雷达，当下一个范围的左界大于雷达安放的坐标，则需要再增加一个雷达在这个范围的右界。若下一个范围的右界大于雷达安放的坐标，则将该雷达移动至右界，数量不变，若小于雷达安放的坐标，则雷达的位置和数量都不变，继续下一个范围的判断。

注意：一定一定要注意用double。

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


const int maxn = 2500;
int ans = 0;

typedef struct NODE
{
    double l;
    double r;

}node;

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

void solve(int n,node node[])
{
    double sign = -1000000000;
    int i;
    for(i = 0;i < n;i++)
    {
        if(node[i].l > sign)
        {
            ans++;
            sign = node[i].r;
        }
        else
        {
            if(node[i].r < sign)
            {
                sign = node[i].r;
            }
        }

    }
}

int main()
{
    node node[maxn];
    int n,d;
    int flag;
    int i;
    int count = 0;
    double x[maxn],y[maxn];
    while(scanf("%d%d",&n,&d) != EOF)
    {
        if(!n && !d) break;
        ans = 0;
        flag = 1;
        count++;
        for(i = 0;i < n;i++)
        {
            scanf("%lf%lf",&x[i],&y[i]);
            if(y[i] > d)
            {
                flag = 0;
            }
        }
        if(!flag)
        {
            printf("Case %d: %d\n",count,-1);
        }
        else
        {
            double z;
            for(i = 0;i < n;i++)
            {
                z = sqrt(d * d * 1.0 - y[i] * y[i]);
                node[i].l = x[i] - z;
                node[i].r = x[i] + z;
            }
            sort(node,node + n,cmp);
            solve(n,node);
            printf("Case %d: %d\n",count,ans);

        }
    }
    return 0;
}