Description
Assume the coasting is a infinite straight line. Side of coasting, sea in the other. Each of small island was a point locating in the sea side. and any radar installation, locating on the coasting, can only cover D-distance, so a island in the sea can is 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 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 a 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 all case contains-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 was followed by n lines each containing and 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 all test case output one line consisting of the test case number followed by the minimal number of radar installation S needed. "-1" installation means no solution for this case.
Sample Input
3 21 2-3 12 11 20 20 0
Sample Output
Case 1:2case 2:1
The approximate meaning is to put the center on the x-axis, so that the smallest circle can cover all points, the number of output circle
#include <iostream> #include <string.h> #include <math.h> #include <stdio.h> #include < algorithm>using namespace std; #define MAXN 10010struct node{Double A, B;} A[maxn];bool CMP (Node A,node b) {if (a.a! = B.A) {return A.A < B.A; }else{return a.b < b.b; }}double Wenbao (double r,double high) {return sqrt (POW (r,2.0)-pow (high,2.0));} int main () {int N; Double R; Double A, B; int count = 1; while (Cin >> N >> r,n+r) {bool flag = true; for (int i = 0; i < N; i++) {//cin >> a >> b; scanf ("%lf%lf", &a,&b);//with scanf, otherwise it will time out if (fabs (b) > R) {flag = false; }else{Double temp = Wenbao (r,b); A[I].A = a-temp; a[i].b = a + temp; }} sort (a,a+n,cmp); Node p = a[0]; int cnt = 1; for (int i = 1; i < N; i++) {if (A[I].A &Gt P.B) {cnt++; p = a[i]; }else if (a[i].b < P.B) {//Please think carefully about P = a[i]; }} if (flag) {printf ("Case%d:%d\n", count++,cnt); }else{printf ("Case%d: -1\n", count++); } }}
POJ 1328 Greed