[POJ1328] Radar Installation

Posted 2020-10-06 HLX_Y

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

 

题解：贪心

按x从小到大排序

首先是要卡边界

对于当前点，如果能被之前的覆盖就直接覆盖

如果不能，那么如果覆盖当前点的圆心（能覆盖到的最右端）的横坐标小于当前圆心横坐标，那么就更新当前圆心坐标，否则ans++

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 #define ll long long
 8 using namespace std;
 9 
10 const int N = 1010;
11 
12 int n,d,ans,tot,flg;
13 double pos;
14 
15 struct Node {
16   double x,y;
17   bool operator < (const Node a) const {
18     return x<a.x;
19   }
20 }p[N];
21 
22 int gi() {
23   int x=0,o=1; char ch=getchar();
24   while(ch!=‘-‘ && (ch<‘0‘ || ch>‘9‘)) ch=getchar();
25   if(ch==‘-‘) o=-1,ch=getchar();
26   while(ch>=‘0‘ && ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
27   return o*x;
28 }
29 
30 int main() {
31   while(scanf("%d%d", &n, &d) && n+d) {
32     ans=1,tot++,flg=0;
33     for(int i=1; i<=n; i++) {
34       p[i].x=gi(),p[i].y=gi();
35     }
36     for(int i=1; i<=n; i++) {
37       if(p[i].y>d) {flg=1;break;}
38     }
39     if(flg) {printf("Case %d: %d\n",tot,-1);continue;}
40     sort(p+1,p+n+1);
41     pos=p[1].x+sqrt(d*d-p[1].y*p[1].y);
42     for(int i=2; i<=n; i++) {
43       if((pos-p[i].x)*(pos-p[i].x)+p[i].y*p[i].y<=d*d) continue;
44       double pos1=p[i].x+sqrt(d*d-p[i].y*p[i].y);
45       if(pos1>pos) ans++;
46       pos=pos1;
47     }
48     printf("Case %d: %d\n",tot,ans);
49   }
50   return 0;
51 }