POJ-1328 Radar Installation

G - Radar Installation

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu

Submit Status

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

AC代码：

#include <stdio.h>  #include <math.h>  #include <algorithm>   using namespace std; #define G 1011 struct node  {      double l,r;  //当小岛被覆盖时，雷达活动的区间 }a[G];  bool cmp(node A,node B)  {      if(A.r!=B.r)    return A.r<B.r;    else    return A.l>B.l; }  int main()  {      int n,d,i,x,y,flag,count;      int k=1;      double temp;      while(scanf("%d%d",&n,&d))      {          if(n==0&&d==0) break;          flag=0;                 //标记是否有小岛无法被覆盖          for(i=0;i<n;++i)          {              scanf("%d%d",&x,&y);              if(y>d)              {                  flag=1;         //如果雷达不足以覆盖，则标记                   continue;               }              temp=sqrt(d*d*1.0-y*y);              a[i].l=x-temp;              a[i].r=x+temp;          }          printf("Case %d: ",k++);          if(flag)          {              printf("-1\n");              continue;          }          sort(a,a+n,cmp);          count = 1;          temp=a[0].r;          for(i=1;i<n;++i)          {              if(a[i].l>temp)              {                  temp=a[i].r;                  count++;              }          }          printf("%d\n",count);      }      return 0;  }