GDUFE ACM-1012-Radar Installation(贪心)

Radar Installation(贪心)

Time Limit: 1000ms

Problem 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.

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:

Sample Output:

Case 1: 2
Case 2: 1

Source:

Beijing 2002


题意：

有一个坐标轴，在x轴的上方时海，下方是陆地，x轴为海岸线，海上有n个小岛，因为种种原因要在海岸线上装雷达，每个雷达的覆盖范围是以d为半径的圆，请用最少的雷达数覆盖所有的小岛，当无法完全覆盖时输出-1；
网上有两种方法：

方法一：将每个区间按照右端点排序，每次在最右段设置雷达，下一个小岛左端点在此雷达左边的点都可以被覆盖，若不满足则设置下一个雷达，向后遍历；

http://blog.csdn.NET/petercsj/article/details/4602946

方法二：按左端点排序，在右端点处设置雷达，若下个点的左端点在此雷达点的右侧，设置新的雷达，向下遍历；若下个点的左端点和右端点均在此雷达点的左边，则不能覆盖，将雷达点位置改变为该点右端点处，此时能够同时覆盖两个点，向下遍历；

http://www.cnblogs.com/zjerly/archive/2011/10/31/2229871.html

思路：我用方法二，

以小岛为圆心以d为半径做圆，计算出圆与x轴的左交点和右交点。左交点为x-sqrt（r*r-y*y），右交点为x+sqrt（r*r+y*y）；然后对左交点从小到大排序，令初始雷达为左交点最小岛屿的右交点，如果i点的左交点在雷达的右面，则需要重新装一个雷达，然后令当前点为新雷达的右交点，否则如果i点的右交点在当前雷达的左边，则把当前雷达的位置更新为该点的右交点。

如图所示



AC代码：

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <algorithm>
 6 using namespace std;
 7 
 8 struct Radar
 9 {
10     double left;
11     double right;
12 }point[1010];
13 inline bool cmp(Radar a,Radar b)
14 {
15     return a.left<b.left;
16 }
17 int main()
18 {
19     int n,r;
20     double x,y;
21     int count=1;
22     while(~scanf("%d%d",&n,&r))
23     {
24         if(n==0&&r==0)
25             break;
26         int ans=1;//雷达的个数
27         for(int i=0;i<n;i++)
28         {
29             scanf("%lf%lf",&x,&y);
30             point[i].left=x-sqrt(r*r-y*y);//当前点与x轴的左交点
31             point[i].right=x+sqrt(r*r-y*y);//当前点与x轴的右交点
32             if(y>r||y<0||r<=0)//列举出三种无法覆盖的情况
33                 ans=-1;
34         }
35         sort(point,point+n,cmp);//左交点排序
36         if(ans!=-1){
37             int j=0;
38             for(int i=1;i<n;i++)
39             {
40                 if(point[i].left>point[j].right)//point[j]为左交点最小岛屿的右交点
41                 {
42                     ans++;
43                     j=i;
44                 }
45                 if(point[i].right<point[j].right)
46                 {
47                     j=i;
48                 }
49             }
50             printf("Case %d: %d
",count++,ans);
51         }
52         else
53             printf("Case %d: -1
",count++);
54     }
55     return 0;
56 }