NYOJ287Radar

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

样例输出

Case 1: 2
Case 2: 1

View Code

 1  
 2 #include<stdio.h>
 3 #include<math.h>
 4 #include<string.h>
 5 #include<algorithm>
 6 using namespace std;
 7 struct dy
 8 {
 9     double left;//**左交点**//
10     double right;//**右交点**//
11 }w[1001];
12 bool comp(dy a,dy b)//**按左交点坐标从小到大排序**//
13 {
14     if(a.left<b.left) return true;
15     return false;
16 }
17 int main()
18 {
19     int n,r,x,y,i,count,num=1;
20     double len,t;
21     while(~scanf("%d %d",&n,&r)&&(n,r))
22     {
23         memset(w,0,sizeof(w));
24         count=1;//**从第一个点开始，所以计数器初值为1**//
25         for(i=0;i<=n-1;i++)
26         {
27             scanf("%d %d",&x,&y);
28             len=sqrt(((double)r*r-(double)y*y));//**结合图形，勾股定理**//
29             w[i].left=(double)x-len;//**左交点的坐标**//
30             w[i].right=(double)x+len;//**右交点的坐标**//
31         }
32         for(i=0;i<=n-1;i++)
33         {
34             if(y>r)//**如果不能完全覆盖**//
35             {
36                 printf("Case %d: -1\n",num++);
37                 break;
38             }
39         }
40         sort(w,w+n,comp);
41         t=w[0].right;
42         for(i=1;i<=n-1;i++)
43         {
44             if(w[i].left>t)//**如果后一个点的左交点大于前一个点的右坐标，说明两点没有公共区域**//
45             {
46                 count++;
47                 t=w[i].right;
48             }
49             else
50             {
51                 if(w[i].right<t)//**如果后一个点的右交点小于前一个点的右坐标，说明后一个点的覆盖区域被前一个点包含了**//
52                 {
53                     t=w[i].right;//**保证后一个点被覆盖**//
54                 }
55             }
56         }
57         printf("Case %d: %d\n",num++,count);
58     }
59     return 0;
60 }