| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 60277 | Accepted: 13583 |
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使用贪心算法解决。
1 #include <iostream>
2 #include<math.h>
3 #include <algorithm>
4 #include <stdlib.h>
5 using namespace std;
6 struct point
7 {
8 double left, right;
9 }points[1000], temp;
10
11 bool operator < (point a, point b)
12 {
13 return a.left < b.left;
14 }
15 int main() {
16 int num = 0;
17 double rad = 0;
18 cin >> num >> rad;
19 int casenum=1;
20 while (num || rad) {
21
22
23 int rnum=1;
24 int wrong = 0;
25
26 for (int i = 0; i < num; i++) {
27 double x=0;
28 double y=0;
29 cin >> x >> y;
30
31 if (y > rad) {
32 //岛屿纵坐标大于雷达半径
33 wrong = 1;
34 }else{
35 //以岛屿为圆心，做圆，与x轴的交点
36 points[i].left=x-sqrt((rad*rad-y*y)*1.0);
37 points[i].right=x+sqrt((rad*rad-y*y)*1.0);
38
39 }
40 }
41 if (wrong) {
42 //雷达不能覆盖
43 cout<<"Case "<<casenum++<<": "<<-1<<endl;
44 } else {
45 //按照岛屿画圆与x轴的左交点排序
46 sort(points, points + num);
47 temp=points[0];
48 for(int i=1;i<num;i++){
49 //
50 if(points[i].left>temp.right)//temp.right为公共区间的右端点，若下一个区间的左端点大于temp.right则该区间与以上的公共区间没有公共部分
51 {
52 rnum++;//雷达个数加1
53 temp=points[i];//更新公共区间右端点
54 }else if(points[i].right<temp.right){
55 temp=points[i];//更新公共区间右端点
56 }
57 }
58
59 cout<<"Case "<<casenum++<<": "<<rnum<<endl;
60 }
61
62 cin >> num >> rad;
63 }
64 return 0;
65 }