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poj 1328 Radar Installation(贪心+快排)

时间:2015-09-30 19:43:22      阅读:220      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

题意:将一条海岸钱看为X轴,X轴的上方为大海,海上有许多岛屿,给出岛屿的位置与雷达的覆盖半径,要求在海岸线上建雷达,
   在雷达能够覆盖所有岛的基础上,求最少需要多少雷达。
技术分享
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <algorithm>
 5 #include <cstring>
 6 #include <cstdlib>
 7 using namespace std;
 8 #define MAX 1005
 9 struct sea
10 {
11     double left;
12     double right;
13 } a[1005];
14 bool operator < (sea A,sea B)
15 {
16     return A.left<B.left;
17 }
18 int main()
19 {
20     int n,k=1;
21     double d;
22     while(cin>>n>>d&&(n||d))
23     {
24         bool flag=false;
25         for(int i=0; i<n; i++)
26         {
27             double x,y;
28             cin>>x>>y;
29             if(fabs(y)>d)
30                 flag=true;
31             else
32             {  //计算区间
33                 a[i].left=x*1.0-sqrt(d*d-y*y);
34                 a[i].right=x*1.0+sqrt(d*d-y*y);
35             }
36         }
37         printf("Case %d: ",k++);
38         if(flag)
39             printf("-1\n");
40         else
41         {
42             int ans=1; //雷达初始化
43             sort(a,a+n); //  排序
44             double s=a[0].right;
45             for(int i=1; i<n; i++)
46             {
47                 if(a[i].left>s)
48                 {
49                     ans++;    //雷达加一
50                     s=a[i].right; // 更新右端点
51                 }
52                 else if(a[i].right<s)
53                     s=a[i].right;
54             }
55             printf("%d\n",ans);
56         }
57     }
58     return 0;
59 }
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poj 1328 Radar Installation(贪心+快排)

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原文地址:http://www.cnblogs.com/cxbky/p/4850097.html

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