贪心poj1328:雷达设置

Posted 2020-10-23 听风不成泣

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

解题思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计 算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内斗至少存在一个点。每次迭代对于第一个区间，选 择最右边一个点， 因为它可以让较多区间得到满足， 如果不选择第一个区间最右一个点（选择前面的点）， 那么把它换成最右的点之后，以前得到满足的区间， 现在仍然 得到满足， 所以第一个区间的最右一个点为贪婪选择， 选择该点之后， 将得到满足的区间删掉，进行下一步迭代， 直到结束。

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <math.h>
 6 using namespace std;
 7 struct node
 8 {
 9     double l,r;
10 };
11 node a[1005];
12 bool cmp(node a,node b)
13 {
14     return a.r<b.r;
15 }
16 int main()
17 {
18     int n,k=1;
19     int flag;
20     double d;
21     while(cin>>n>>d)
22     {
23         flag=1;
24         if(n==0&&d==0)
25         {
26             break;
27         }
28         double x,y;
29         for(int i=0;i<n;i++)
30         {
31             cin>>x>>y;
32             if(!flag)
33             {
34                 continue;
35             }
36             if(y>d)
37             {
38                 flag=0;
39                 continue;
40             }
41             a[i].l=x-sqrt(d*d-y*y);
42             a[i].r=x+sqrt(d*d-y*y);
43         }
44         printf("Case %d: ", k++);
45         if(!flag)
46         {
47             printf("-1\n");
48             continue;
49         }
50         sort(a,a+n,cmp);
51         double temp=-10000;
52         int ans=0;
53         for(int i=0;i<n;i++)
54         {
55             if(temp<a[i].l)
56             {
57                 ans++;
58                 temp=a[i].r;
59             }
60         }
61         cout << ans << endl;
62     }
63 
64     return 0;
65 }

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