poj 1328Radar Installation

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Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 53901 Accepted: 12143

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 <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;double x,y;struct point{    double  x1;    double  x2;} p[1001];int cmp(point a,point b){    return a.x1<b.x1;}int main(){    int n,d;    //int ans=1;    int kase=0;    while(~scanf("%d%d",&n,&d)&&n+d)    {        int ans=1;        for(int i=0; i<n; i++)        {            cin>>x>>y;            p[i].x1=x-sqrt(d*d-y*y);            p[i].x2=x+sqrt(d*d-y*y);            if(y>d||y<0||d<=0)                ans=-1;        }        sort(p,p+n,cmp);        double xx=p[0].x2;        for(int i=1; i<n && ans!=-1; i++)        {            if(p[i].x1>xx)            {                ans++;                xx=p[i].x2;            }            else if(p[i].x2<xx)                {                    xx=p[i].x2;                }        }        printf("Case %d: %d\n",++kase,ans);    }}