POJ 1328 Radar Installation
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Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 49398 Accepted: 11037
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
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
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
#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#include <ctype.h>#include<algorithm>using namespace std;struct node{ double x,y;} a[1001];bool cmp(node x,node y){ return x.x<y.x;}int main(){ int n,d,i,j,x,y,l=0; while(scanf("%d%d",&n,&d)!=EOF&&(n!=0&&d!=0)) { l++; int flag=0; for(i=0; i<n; i++) { scanf("%d%d",&x,&y); if(y>d) flag=1; a[i].x=x-sqrt(d*d-y*y); a[i].y=x+sqrt(d*d-y*y); } if(flag) { printf("Case %d: -1\n",l); continue; } sort(a,a+n,cmp); int ans=1; double border=a[0].y; for(i=0; i<n; i++) { if(a[i].x>border) { ans++; border=a[i].y; } if(a[i].y<border) border=a[i].y; } printf("Case %d: %d\n",l,ans); } return 0;}
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