C题

C - C

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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<tex2html_verbatim_mark> distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d<tex2html_verbatim_mark> .

We use Cartesian coordinate system, defining the coasting is the x<tex2html_verbatim_mark> -axis. The sea side is above x<tex2html_verbatim_mark> -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<tex2html_verbatim_mark> - y<tex2html_verbatim_mark>coordinates.

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Input

The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1n1000)<tex2html_verbatim_mark> and d<tex2html_verbatim_mark> , where n<tex2html_verbatim_mark> is the number of islands in the sea and d<tex2html_verbatim_mark> is the distance of coverage of the radar installation. This is followed by n<tex2html_verbatim_mark> 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

题解：求给出的点，雷达最少有几个才能完全覆盖。

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
    double l,r;
};
node a[];
int cmp(node a,node b)
{
    return a.l<b.l;
}
int main()
{
    int n,d;
    int x,y,f,t;
    int s=;
    while(cin>>n>>d&&n&&d)
    {
        f=;
        t=;
        s++;
        for(int i=; i<n; i++)
        {
            cin>>x>>y;
            if(y>d)f=;
            else
            {
                a[i].l=(double)x-sqrt((double)d*d-y*y);
                a[i].r=(double)x+sqrt((double)d*d-y*y);
            }
        }
        if(f==)
        {
            cout<<"Case "<<s<<": -1"<<endl;
            continue;
        }
        sort(a,a+n,cmp);
        double p=a[].r;
        for(int i=; i<n; i++)
        {

            if(a[i].r<p)
                p=a[i].r;
            else if(p<a[i].l)
            {
                p=a[i].r;
                t++;
            }
        }
        cout<<"Case "<<s<<": "<<t<<endl;

    }
    return ;
}