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贪心 Radar Installation (求最少探测雷达)

时间:2016-07-23 23:03:01      阅读:225      评论:0      收藏:0      [点我收藏+]

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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. 
技术分享
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



先算出每个岛放探测雷达的x坐标范围,b为雷达探测半径,岛的坐标为(x,y),雷达的坐标范围是(x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)),定义sum=1,i从1开始,每个坐标范围左右边界与上一个坐标范围的右边界比较,具体看代码。
 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cmath>
 4 using namespace std;
 5 struct stu
 6 {
 7     double l,r;
 8 } st[1500];
 9 bool cmp(stu a,stu b)
10 {
11     return a.l<b.l;
12 }
13 int main()
14 {
15     int a,s,k=0,sum,i;
16     double b,x,y,sky;
17     while(scanf("%d %lf",&a,&b) &&!(a==0&&b==0))
18     {
19         s=0;
20         k++;
21         for(i = 0 ; i < a ; i++)
22         {
23             scanf("%lf %lf",&x,&y);
24             if(b < 0 || y > b)
25             {
26                 s=-1;
27             }
28             st[i].l=x-sqrt(b*b-y*y);
29             st[i].r=x+sqrt(b*b-y*y);
30         }
31         if(s == -1) printf("Case %d: -1\n",k);
32         else
33         {
34             sort(st,st+a,cmp);
35             sum=1;
36             sky=st[0].r;
37             for(i = 1 ; i < a ; i++)
38             {
39                 if(st[i].r <= sky)
40                 {
41                     sky=st[i].r;
42                 }
43                 else
44                 {
45                     if(st[i].l > sky)
46                     {
47                         sky=st[i].r;
48                         sum++;
49                     }
50                 }
51             }
52             printf("Case %d: %d\n",k,sum);
53         }
54         
55     }
56 }

 

贪心 Radar Installation (求最少探测雷达)

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原文地址:http://www.cnblogs.com/yexiaozi/p/5699559.html

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