反正每一个岛都要有雷达覆盖,放雷达的时候尽可能多覆盖岛......
#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 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.
Input
The input consists of several test cases. The first line of each case contains two integers
n (1n1000)
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 2 1 2 -3 1 2 1 1 2 0 2 0 0
Sample Output
Case 1: 2 Case 2: 1