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POJ 1328 Radar lnstallation 2

2024-09-09 07:26:11   218人阅读

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 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1







贪心：注意重点
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
int f[1010];
struct aa
{
    double x,y;
}a[1010];
int cmp(aa q,aa w)
{
    return q.x<w.x;
}
int main()
{
    int k,n,m,cnt,ka=0;
    while(cin>>n>>m)
    {   if(n==0&&m==0)break;
        ka++;
        int flat=1;
        k=0;
        double x,y,c,mm=(double)m*m;
        for(int i=1;i<=n;i++)
        {
          cin>>x>>y;
          if(m>=y)
          {c=sqrt(mm-y*y);
           a[++k].x=x-c;
           a[k].y=x+c;
          }
          else
         {
             flat=0;
          }

        }
        getchar();
        if(flat)
        {sort(a+1,a+k+1,cmp);
        double now = a[1].y;
         cnt = 1;
        for(int i=2;i<=k;i++)
        {
          if(a[i].y<now)
            now=a[i].y;
          else if(a[i].x>now)
          {
              cnt++;
              now=a[i].y;
          }
        }
        }
        else
        cnt=-1;
        cout<<"Case "<<ka<<": "<<cnt<<endl;
    }
    return 0;
}

POJ 1328 Radar lnstallation 2


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