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poj 1328 Radar Installation

2024-08-05 18:23:44   222人阅读

http://poj.org/problem?id=1328

贪心：

题意：给定小岛的位置，问最小安装雷达数，雷达有一个半径，覆盖小岛。我竟然看错了题意，雷达都安装在x轴上，我竟然不知道。

分析：以小岛做一个圆，圆与x轴有2个交点a,b在区间[a,b]选点的问题，就转换区间选点问题，在小白书1的153页，先把区间b从小到大排序，取最后一个点。

                                                                                       Radar Installation
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 55095 Accepted: 12415

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1
 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 struct node 8 { 9     double x,y;10 }a[1005];11 bool cmp(node d,node c)12 {13     return d.y<c.y;14 }15 int main()16 {17     int n,i,j,t=1,flag,ans;18     double d,r,w;19     while(~scanf("%d%lf",&n,&r))20     {21             ans=1;22            flag=0;23         if(n==0&&r==0)24              break;25         for(i=0;i<n;i++)26         {27             scanf("%lf%lf",&w,&d);28             if((r*r-d*d)<0)29                 flag=1;30              a[i].x=w-sqrt(r*r-d*d);31               a[i].y=w+sqrt(r*r-d*d);32         }33           if(flag||r<=0)34            {35                 printf("Case %d: -1\n",t++);36                 continue;37            }38         sort(a,a+n,cmp);39         double temp=a[0].y;40         for(i=1;i<n;i++)41         {42             if(a[i].x>temp)43               {44                   ans++;45                   temp=a[i].y;46               }47         }48         printf("Case %d: %d\n",t++,ans);49     }50     return n;51 }

 

poj 1328 Radar Installation


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