can space des math pst sep 當前 nds not

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

題解：貪心

按x從小到大排序

首先是要卡邊界

對於當前點，如果能被之前的覆蓋就直接覆蓋

如果不能，那麽如果覆蓋當前點的圓心（能覆蓋到的最右端）的橫坐標小於當前圓心橫坐標，那麽就更新當前圓心坐標，否則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 }

[POJ1328] Radar Installation