POJ1328 -- Radar Installation

Radar Installation
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 96747   Accepted: 21508

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

Source

Beijing 2002
 

题意分析
给定一个直角坐标系，定义x轴为海岸线，海岸线上方是海，下方是陆地.
在海域零星分布一些海岛, 需要要在海岸线上安装若干个雷达覆盖这些海岛,
每个雷达的覆盖半径都是相同且固定的.

现在给定所有海岛的坐标(x,y), 以及雷达的覆盖半径d，
问可以覆盖所有海岛的最小雷达数.

 


解题思路：

无解的情况：

　　y>d或者是y<0或是d<0

 

有解的情况：

以海岛坐标(x,y)为圆心，用雷达覆盖半径d画圆，根据前提条件可知，
这个圆与x轴必定存在最少1个(y=d)、最多2个交点(y<d).

可以认为1个交点是2个交点重合的特殊情况，那么这2个交点在x轴上构成的线性区间，
可以看作海岛的被覆盖范围在x轴上的投影 (由此就可以把二维的几何问题转化成一维几何问题)

按照所有海岛的x轴坐标，从小到大依次计算每个海岛在x轴上的投影区间（投影可能存在重叠），
同时把每个雷达抽象成1个点，那么此问题就转化成：

【已知x轴上若干个区间（可能存在交集），现在要往这些区间内放若干个点，
问怎样放置这些点，使得每个区间内至少存在一个点，且所放置的点的总量尽可能最少】


可使用贪心算法求解：
根据每个区间的左端点坐标对所有区间从左到右排序:
① 在左起第一个区间A 的右端点 A.R 放置一个点，总放置点数 P+1
② 若下一个区间B 的左端点 B.L > A.R,
说明 区间A 与 区间B 无交集，此时在区间B 的右端点 B.R 放置一个点，总放置点数 P+1

否则说明 区间A 与 区间B 相交, 此时进一步判断，若 B.R < A.R，
说明 区间B 在 区间A 的内部，此时把原本放置在 A.R 的点移动到 B.R（确保区间B有一个点），总放置点数不变

③ 重复这个过程直到所有区间被遍历完成

 

测试样例：
3 2
1 0
-3 -1
2 1

2 2
1 1
2 2

5 4
-2 2
1 2
0 3
3 3
4 3

9 3
0 2
-3 2
-4 2
-5 2
3 2
6 2
9 2
12 2
15 2

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

 

 

 1 #include<iostream>
 2 #include<math.h>
 3 #include<algorithm>
 4 using namespace std;
 5 struct seaside{
 6     double left;
 7     double right;
 8 };
 9 bool compare(seaside a,seaside b)
10 {
11     return a.left<=b.left;
12 }
13 int solve(seaside *a,int n)
14 {
15     ///进行排序
16     sort(a,a+n,compare);
17     double fright;
18     fright = a[0].right;
19     int sum = 1;
20     for(int i=1;i<n;i++)
21     {
22         if(a[i].left > fright)
23         {
24             sum++;
25             fright = a[i].right;
26         }else{
27             if(a[i].right < fright)
28             {
29                 fright = a[i].right;
30             }
31         }
32     }
33     return sum;
34 }
35 int main()
36 {
37     int n;
38     double d;
39     int Count = 1;
40     bool isSolved = 1;
41     cin>>n>>d;
42     while(n!=0)
43     {
44         isSolved = 1;
45         seaside *island = new seaside[n];
46         double x,y;
47         for(int i=0;i<n;i++)
48         {
49             cin>>x>>y;
50             if(y>d || y<0 || d<0)//无解
51                 isSolved = 0;
52             else{
53                 double offset = sqrt(d*d - y*y);   // 勾股定理
54                 island[i].left = x-offset;
55                 island[i].right = x+offset;
56             }
57 
58         }
59 
60         ///解决问题
61         cout<<"Case "<<Count<<": ";
62         if(isSolved)
63             cout<<solve(island,n)<<endl;
64         else
65             cout<<-1<<endl;
66 
67         delete[] island;
68         Count++;
69         cin>>n>>d;
70     }
71     return 0;
72 }

 

 

结果：
3 2
1 0
-3 -1
2 1
Case 1: -1

2 2
1 1
2 2
Case 2: 1

5 4
-2 2
1 2
0 3
3 3
4 3
Case 3: 1

9 3
0 2
-3 2
-4 2
-5 2
3 2
6 2
9 2
12 2
15 2
Case 4: 4

3 2
1 2
-3 1
2 1
Case 5: 2

1 2
0 2
Case 6: 1

0 0

Process returned 0 (0x0)   execution time : 1.058 s
Press any key to continue.

 

posted @ 2018-02-03 16:23  卉卉卉大爷  阅读(257)  评论(0)  编辑  收藏  举报