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 }