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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Input
The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1
n1000)<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.
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 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。