Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 60120 Accepted: 13552

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1
#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <cstdlib>#include <cctype>#include <queue>#include <stack>#include <algorithm>using namespace std;struct node//记录每个岛的安装雷达的范围{    double L;    double R;}point[1100];bool cmp(node a,node b)//sort比较函数{    return a.L<b.L;}int main(){    int n,d;    int x,y;    int w=1;    bool flag;    while(scanf("%d %d",&n,&d))    {        if(n==0&&d==0)        {            break;        }        int top=0;        flag=true;        for(int i=0;i<n;i++)        {            scanf("%d %d",&x,&y);            if(d<y)//如果有不符合的记录            flag=false;            if(flag)            {                point[top].L=x-sqrt(d*d-y*y);                point[top].R=x+sqrt(d*d-y*y);                top++;            }        }        printf("Case %d: ",w++);        if(flag)        {            sort(point,point+top,cmp);            double ans=point[0].R;            int sum=1;            for(int i=1;i<top;i++)            {                if(point[i].L>ans)//如果按装的范围不能包括，就增加雷达的个数                {                    ans=point[i].R;                    sum++;                }                else if(point[i].R<ans)                {                    ans=point[i].R;//雷达范围的更新                }            }            printf("%d\n",sum);        }        else        {            printf("-1\n");        }    }    return 0;}