Radar Installation
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 54 Accepted Submission(s) : 28
Problem 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
题解：先转化为区间点，再排序，区间找点；
代码：
1 #include<stdio.h> 2 #include<math.h> 3 #include<algorithm> 4 using namespace std; 5 struct Node{ 6 double s,e; 7 }; 8 int n,d,k; 9 Node area[1010]; 10 int cmp(Node a,Node b){ 11 return a.e<b.e; 12 } 13 int change(int x,int y){ 14 if(y>d)return 0; 15 double a,b,m=sqrt(d*d-y*y); 16 a=x-m;b=x+m; 17 area[k].s=a;area[k].e=b;k++; 18 return 1; 19 } 20 int main(){int t,x,y,flot,temp,num,l=0; 21 while(scanf("%d%d",&n,&d),n||d){k=0;flot=1;temp=0;num=1;l++; 22 for(int i=0;i<n;i++){ 23 scanf("%d%d",&x,&y); 24 t=change(x,y); 25 if(!t)flot=0; 26 } 27 sort(area,area+k,cmp); 28 for(int i=0;i<k;i++){ 29 if(area[i].s>area[temp].e)temp=i,num++; 30 } 31 printf("Case %d: %d ",l,num); 32 } 33 return 0; 34 }