Radar coverage is required for every island. When radar is used, cover as many islands as possible ......
#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>using namespace std;struct node{ double x,y;};bool cmp(node a,node b){ return a.x<b.x;}int main(){ // freopen("in","r",stdin); int T,i,n,ans,jishu=0; double d,pr,l,r,t; node box[1010]; while(cin>>n>>d) { if(n==0&&d==0) break; for(i=0;i<n;i++) cin>>box[i].x>>box[i].y; sort(box,box+n,cmp); t=sqrt((d+box[0].y)*(d-box[0].y)); pr=box[0].x+t; ans=1; for(i=0;i<n;i++) { if(d<fabs(box[i].y)) { ans=-1; break; } t=sqrt((d+box[i].y)*(d-box[i].y)); l=box[i].x-t; r=box[i].x+t; if(l>pr) { pr=r; ans++; } else if(r<pr) pr=r; } printf("Case %d: %d\n",++jishu,ans); }}
D.
We use Cartesian coordinate system, defining the coasting isX-Axis. The sea side is abveX-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 itsX-YCoordinates.
Input
The input consists of several test cases. The first line of each case contains two integersN(1N1000) andD, WhereNIs the number of islands in the sea andDIs the distance of coverage of the radar installation. This is followedNLines 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
Uva-2519-radar installation