poj-1328-Radar Installation

向右看齐 2022-05-20 07:06 157阅读 0赞

Radar Installation

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;104992</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;23304</td> 
  </tr> 
 </tbody> 
</table>

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.   
![1328_1.jpg][]   
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

Source

[Beijing 2002][]

题目大意：在一个二维直角坐标系中，给出n个海岛的坐标，现在在海岸线上装雷达，给出雷达的覆盖半径，问如何布置雷达，才能使雷达的数量最少，并求出最少的雷达数量

思路：典型的贪心，我们可以在输入时对其判断，若输入的坐标y最大的就大于半径，那不管如何布置，都将不能覆盖，现在我们不妨设每个海盗就是这个圆的圆心，求其与海岸线的交点，这个交点就是这个圆能覆盖的最远距离，然后将海岛变成区间，然后用区间覆盖解法去做就可以了

代码：

#include<map>
    #include<stack>
    #include<cmath>
    #include<string>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define maxn 1100
    #define ll long long
    struct qq
    {
        double l,r;
    } a[maxn];
    int cmp(qq A,qq B)
    {
        return A.r<B.r;
    }
    int main()
    {
        int d,n,q=1;
        while(scanf("%d%d",&n,&d)!=EOF)
        {
            if(d==0&&n==0)
                break;
            int i,j,k,x,y,ans=0,flag=0;
            double dis;
            for(i=0; i<n; i++)
            {
                scanf("%d%d",&x,&y);
                dis=sqrt((double)(d*d)-(double)(y*y));
                a[i].l=(double)x-dis;
                a[i].r=(double)x+dis;
                if(y>d)
                    flag=1;
            }
            sort(a,a+n,cmp);
            printf("Case %d: ",q++);
            if(flag)
            {
                printf("-1\n");
                continue;
            }
            double temp=-99999999;
            for(i=0; i<n; i++)
            {
                if(a[i].l>temp)
                {
                    ans++;
                    temp=a[i].r;
                }
            }
            printf("%d\n",ans);
        }
        return 0;
    }

[1328_1.jpg]: /images/20220520/cdc29ca671df46f5bcffa86568e1d344.png
[Beijing 2002]: http://poj.org/searchproblem?field=source&key=Beijing+2002