Radar Installation 分类： POJ 2015-06-15 19:54 8人阅读 评论(0) 收藏

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 60120		Accepted: 13552

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 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
struct node//记录每个岛的安装雷达的范围
{
    double L;
    double R;
}point[1100];
bool cmp(node a,node b)//sort比较函数
{
    return a.L<b.L;
}
int main()
{
    int n,d;
    int x,y;
    int w=1;
    bool flag;
    while(scanf("%d %d",&n,&d))
    {
        if(n==0&&d==0)
        {
            break;
        }
        int top=0;
        flag=true;
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&x,&y);
            if(d<y)//如果有不符合的记录
            flag=false;
            if(flag)
            {
                point[top].L=x-sqrt(d*d-y*y);
                point[top].R=x+sqrt(d*d-y*y);
                top++;
            }
        }
        printf("Case %d: ",w++);
        if(flag)
        {
            sort(point,point+top,cmp);
            double ans=point[0].R;
            int sum=1;
            for(int i=1;i<top;i++)
            {
                if(point[i].L>ans)//如果按装的范围不能包括，就增加雷达的个数
                {
                    ans=point[i].R;
                    sum++;
                }
                else if(point[i].R<ans)
                {
                    ans=point[i].R;//雷达范围的更新
                }
            }
            printf("%d\n",sum);
        }
        else
        {
            printf("-1\n");
        }
    }
    return 0;
} 

版权声明：本文为博主原创文章，未经博主允许不得转载。