NYOJ 287 Radar

Radar

时间限制：1000 ms  |           内存限制：65535 KB
难度：3
 
描述
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.

 

 
输入
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
输出
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.
样例输入
3 2
1 2
-3 1
2 1

1 2
0 2

0 0
样例输出
Case 1: 2
Case 2: 1
思路：确定岛在x轴上可覆盖区间（即在此范围内，信号可覆盖岛）右边界排序；
代码：
 1  
 2 #include<stdio.h>
 3 #include<math.h>
 4 #include<algorithm>
 5 using namespace std;
 6 const int MAXN=1005;
 7 struct Line
 8 {
 9     double l,r;
10 }line[MAXN];//每个岛作半径为d的圆，截得区间
11 bool cmp(Line a,Line b)
12 {
13     return a.r<b.r;
14 }
15 int main()
16 {
17     int n,d;
18     int i;
19     int x,y;
20     int num=1,flag,count;
21     while(scanf("%d %d",&n,&d)&&n&&d)
22     {
23         flag=1;
24         for(i=0;i<n;i++)
25         {
26             scanf("%d %d",&x,&y);
27             if(!flag) continue;
28             if(y<=d){
29               line[i].l=(double)x-sqrt((double)d*d-y*y);
30                line[i].r=(double)x+sqrt((double)d*d-y*y);
31             }else flag=0;
32         }
33         if(!flag)
34         {
35             printf("Case %d: -1\n",num++);
36             continue;
37         }
38         sort(line,line+n,cmp);
39         count=1;
40         double now=line[0].r;
41         for(i=1;i<n;i++)
42         {
43 
44             if(line[i].l<=now+0.00005) continue;
45             now=line[i].r;
46             count++;
47         }
48       printf("Case %d: %d\n",num++,count);
49     }
50     return 0;
51 }
52
View Code

 

posted on 2013-08-29 10:47  落水寒冰  阅读(226)  评论(0)  编辑  收藏  举报

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