cas bsp sep 得到 find most air seve 小數

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

解題思路：該題題意是為了求出能夠覆蓋所有島嶼的最小雷達數目，每個小島對應x軸上的一個區間，在這個區間內的任何一個點放置雷達，則可以覆蓋該小島，區間範圍的計 算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];這樣，問題即轉化為已知一定數量的區間，求最小數量的點，使得每個區間內鬥至少存在一個點。每次叠代對於第一個區間，選 擇最右邊一個點， 因為它可以讓較多區間得到滿足， 如果不選擇第一個區間最右一個點（選擇前面的點）， 那麽把它換成最右的點之後，以前得到滿足的區間， 現在仍然 得到滿足， 所以第一個區間的最右一個點為貪婪選擇， 選擇該點之後， 將得到滿足的區間刪掉，進行下一步叠代， 直到結束。

 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 }

View Code

【貪心】poj1328:雷達設置