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.

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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>(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.

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

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 }

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