Radar Installation

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Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 77760 Accepted: 17398

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

Source

Beijing 2002

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<math.h>#include<algorithm>using namespace std;struct node{    double l;    double r;} a[1010];bool comp(node a,node b){    if(a.r<b.r)        return true;    else if(a.r==b.r)    {        if(a.l>b.l)            return true;        return false;    }    return false;}int main(){    int num;    int i,n;    int cas=1;    int x,d,y;    double temp;    while(~scanf("%d %d",&n,&d)&&(n||d))    {        int flag=1;        num=1;        for(i=0; i<n; i++)        {            scanf("%d %d",&x,&y);            temp=(double)(d*d-y*y);            if(temp<0||d<0)            {                flag=0;            }            else            {                a[i].l=x-sqrt((double)d*d-y*y);                a[i].r=x+sqrt((double)d*d-y*y);            }        }        if(!flag)        {            printf("Case %d: -1\n",cas++);            continue;        }        else        {            sort(a,a+n,comp);            temp=a[0].r;            for(i=1; i<n; i++)            {                if(a[i].l>temp)                {                    temp=a[i].r;                    num++;                }            }            printf("Case %d: %d\n",cas++,num);        }    }    return 0;}


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