poj1328 Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 59264 Accepted: 13349

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

题意大概就是，给出几个坐标，问最少的圆能覆盖所有点的数目。还有非法的可能。

贪心吧

#include <iostream>#include <string.h>#include <math.h>#include <algorithm>using namespace std;int n,d;struct point{int x,y;double xx;}p[1010];int flag;bool compare(point a,point b){return a.xx<b.xx;}int vis[1010];int main(){int _case=0;while(cin>>n>>d){flag=0;if(n==0&&d==0) break;for(int i=1;i<=n;i++){cin>>p[i].x>>p[i].y;if(p[i].y>d) flag=1;else{p[i].xx=p[i].x+sqrt(d*d-p[i].y*p[i].y);}}if(flag==1){cout<<"Case "<<++_case<<": ";cout<<"-1"<<endl;continue;}sort(p+1,p+n+1,compare);int cnt=0;memset(vis,0,sizeof(vis));//cout<<"asdf";for(int i=1;i<=n;i++){if(!vis[i]){cnt++;for(int j=1;i+j<=n;j++){if(((p[i].xx-p[i+j].x)*(p[i].xx-p[i+j].x)+p[i+j].y*p[i+j].y)<=d*d){vis[i+j]=1;}else break;}}}cout<<"Case "<<++_case<<": ";cout<<cnt<<endl;}return 0;}