Radar Installation
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 54970 | Accepted: 12381 |
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 2 1 2 -3 1 2 1 1 2 0 2 0 0
Sample Output
Case 1: 2 Case 2: 1
Source
题目翻译:给出两个数字n和d,n代表n个P点,d代表圆的半径,然后下面n行,每行一组坐标(x,y),分别代表所有的P点,求:在x坐标轴上,至少要画几个以d为半径的圆,才能使P点均在圆内。
解题思路:分别以P点为圆心,d为半径画圆,圆分别会在x轴上截取相应的线段,只要线段内有圆心,则即可覆盖P点,因此问题
便转化为求X轴上共有多少独立区间!
#include<iostream> #include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> #include<math.h> using namespace std; struct A{ double zuo,you; }; int cmp(A a,A b){ return a.zuo<b.zuo; } int main(){ int t=1,n,i,r,x,y,yes; A a[1200]; while(scanf("%d%d",&n,&r)&&!(n==0&&r==0)){ yes=0; for(i=1;i<=n;i++){ scanf("%d%d",&x,&y); if(r<y) yes=1; double num=sqrt((double)(r*r-y*y)); a[i].zuo=double(x-num); a[i].you=double(x+num); } if(yes){printf("Case %d: -1\n",t++);continue;} sort(a+1,a+n+1,cmp); double k=a[1].you; int count=1; for(i=2;i<=n;i++){ if(a[i].zuo>k){ k=a[i].you; count++; } else{ if(a[i].you<k) k=a[i].you; } } printf("Case %d: %d\n",t++,count); } return 0; }