Radar Installation(贪心,可以转化为今年暑假不ac类型)
Radar Installation
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 54 Accepted Submission(s) : 28
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
The input is terminated by a line containing pair of zeros
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct Node{
double s,e;
};
int n,d,k;
Node area[];
int cmp(Node a,Node b){
return a.e<b.e;
}
int change(int x,int y){
if(y>d)return ;
double a,b,m=sqrt(d*d-y*y);
a=x-m;b=x+m;
area[k].s=a;area[k].e=b;k++;
return ;
}
int main(){int t,x,y,flot,temp,num,l=;
while(scanf("%d%d",&n,&d),n||d){k=;flot=;temp=;num=;l++;
for(int i=;i<n;i++){
scanf("%d%d",&x,&y);
t=change(x,y);
if(!t)flot=;
}
sort(area,area+k,cmp);
for(int i=;i<k;i++){
if(area[i].s>area[temp].e)temp=i,num++;
}
printf("Case %d: %d\n",l,num);
}
return ;
}
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