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
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
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 }
Radar Installation---（贪心）
原文：https://www.cnblogs.com/Bravewtz/p/10539029.html
