装雷达

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<tex2html_verbatim_mark> distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d<tex2html_verbatim_mark> .

We use Cartesian coordinate system, defining the coasting is the x<tex2html_verbatim_mark> -axis. The sea side is above x<tex2html_verbatim_mark> -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<tex2html_verbatim_mark> - y<tex2html_verbatim_mark>coordinates.

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The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1n1000)<tex2html_verbatim_mark> and d<tex2html_verbatim_mark> , where n<tex2html_verbatim_mark> is the number of islands in the sea and d<tex2html_verbatim_mark> is the distance of coverage of the radar installation. This is followed by n<tex2html_verbatim_mark> 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.

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.

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Case 1: 2
Case 2: 1


思路：
    贪心，找重复区间


源代码：

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<string>
 7 using namespace std;
 8 #define maxn 1000+5
 9 struct node
10 {
11     double left;
12     double right;
13     bool operator<(const node&a)const
14     {
15         return left < a.left;
16     }
17 }line[maxn];
18 int main()
19 {
20     int n, d;
21     int flag, ans=0, k;
22     while (cin >> n >> d)
23     {
24         ans++;
25         flag = 1;
26         k = 0;
27         int x, y;
28         if (n == 0 && d == 0)
29             break;
30         for (int i = 0; i < n; i++)
31         {
32             cin >> x >> y;
33             
34             if (flag == 0)continue;
35             if (y>d)
36                 flag = 0;
37             else
38             {
39                 /*Index = 0;*/
40                 line[i].left = (double)x- sqrt((double)d*d-(double)y * y);
41                 line[i].right = (double)x + sqrt((double)d*d - (double)y * y);
42                 
43             }
44         }
45         if (flag == 0)
46         {
47             cout << "Case " << ans << ": -1" << endl;
48             continue;
49         }
50         sort(line, line + n);
51         k++;
52         double cur = line[0].right;
53         for(int i = 1; i < n; i++)
54         {
55             if (line[i].right < cur)  
56             {
57                 cur = line[i].right;
58 
59             }
60             else if (cur<line[i].left)
61             {
62                 cur = line[i].right;
63                 k++;
64             }
65         
66         }
67         
68         cout << "Case " << ans << ": "<<k << endl;
69     }
70     return 0;
71 }

心得：

   好吧，这个题目自己都还没完全搞懂，也是醉了~~~，好好研究，之后懂了再补上说明。