Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 115873 | Accepted: 25574 |
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
分析：首先根据小岛的坐标计算出每座小岛对应海岸线上的范围。将每个小岛对应在海岸线上的范围进行排序，使得每个雷达范围的最小值进行递增。
对雷达范围进行贪心。。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 using namespace std; 7 const int maxn=1010; 8 #define INf 0x3f3f3f3f 9 struct node{ 10 double l,r; 11 }point[maxn]; 12 13 bool cmp(const node &a,const node &b){ 14 return a.l<b.l; 15 } 16 17 int main(){ 18 int n,d; 19 int case1=0; 20 while(~scanf("%d%d",&n,&d)&&n){ 21 int flag=0; 22 for(int i=0; i<n; i++ ){ 23 int x,y; 24 cin>>x>>y; 25 if(y>d){ 26 flag=1; 27 // break; 28 } 29 double p=sqrt((double)(d*d)-y*y); 30 point[i].l=x-p; 31 point[i].r=x+p; 32 } 33 printf("Case %d: ",++case1); 34 if(flag){ 35 cout<<-1<<endl; 36 continue; 37 } 38 sort(point,point+n,cmp); 39 int ans=1; 40 node tmp=point[0]; 41 for( int i=1; i<n; i++ ){ 42 if(tmp.r>=point[i].r) tmp=point[i]; 43 else if(tmp.r<point[i].l){ 44 ans++; 45 tmp=point[i]; 46 } 47 } 48 cout<<ans<<endl; 49 } 50 return 0; 51 }