C-C Radar Installation 解题报告
题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86640#problem/C
题目：
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<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.
Input
The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1n
1000)<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
题目大意：
在一条海岸线的一侧有若干个岛屿，将海岸线看成是X轴，X轴以上是大海，在海岸上安装雷达使其能覆盖全部岛屿，给出雷达的覆盖半径和岛屿位置，求最少使用多少
雷达才能将所有岛屿全部覆盖。无法覆盖输出-1.
分析：
1.可以换一种想法，求雷达覆盖小岛就是以小岛为圆心，雷达的覆盖半径为半径画圆
2.小岛到海岸线的距离大于半径，不能覆盖所有小岛，输出-1
3.若小于半径，该圆与X轴的左交点为line[i].l=(double)x-sqrt((double)d*d-y*y);右交点为 line[i].r=(double)x+sqrt((double)d*d-y*y);
4.将左交点排序，判断雷达位置，如果i点的左交点在当前雷达的右边，需要安装一个新雷达；如果i点的右交点在当前雷达的左边，则把当前雷达的圆心更新为该点的右交点
代码：
1 #include<cstdio>
2 #include<iostream>
3 #include<cmath>
4 #include<algorithm>
5 using namespace std;
6
7 const int maxn=1005;
8
9 struct Line //每个岛作半径为d的圆，与x轴所截的线段
10 {
11 double l,r;
12 }line[maxn];
13
14 bool cmp(Line a,Line b)//按照线段的左端点从小到大排序
15 {
16 return a.l<b.l;
17 }
18
19 int main()
20 {
21 int n,d;
22 int i;
23 int x,y;
24 bool yes;//确定是不是有解
25 int m=1;
26 while(scanf("%d%d",&n,&d)!=EOF)
27 {
28 yes=true;
29 int cnt=0;
30 if(n==0&&d==0) //n,d均为0，程序结束
31 break;
32 for(i=0;i<n;i++)
33 {
34 scanf("%d%d",&x,&y);
35 if(yes==false)
36 continue;
37 if(y>d)
38 yes=false;
39 else
40 {
41 line[i].l=(double)x-sqrt((double)d*d-y*y);
42 line[i].r=(double)x+sqrt((double)d*d-y*y);
43 }
44 }
45 if(yes==false) //不能完全覆盖
46 {
47 printf("Case %d: -1
",m++);
48 continue;
49 }
50 sort(line,line+n,cmp);//排序
51 cnt++;
52 double now=line[0].r;
53 for(i=1;i<n;i++)
54 {
55 if(line[i].r<now) //与现在比较，这个很重要
56 now=line[i].r;
57 else if(now<line[i].l)
58 {
59 now=line[i].r;
60 cnt++;
61 }
62 }
63 printf("Case %d: %d
",m++,cnt);
64 }
65
66 return 0;
67 }