poj 1328Radar Installation

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最新推荐文章于 2025-03-30 11:00:00 发布

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本文探讨了在无限海岸线上，通过雷达安装实现岛屿覆盖的最小数量问题。使用二维坐标系，通过输入岛屿的位置和雷达的覆盖距离，计算并输出所需雷达安装的数量。通过实例演示了解决此类问题的方法。

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Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 53901		Accepted: 12143

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

Sample Output

Case 1: 2
Case 2: 1

Source

Beijing 2002

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
double x,y;

struct point
{
    double  x1;
    double  x2;
} p[1001];

int cmp(point a,point b)
{
    return a.x1<b.x1;
}

int main()
{
    int n,d;
    //int ans=1;
    int kase=0;
    while(~scanf("%d%d",&n,&d)&&n+d)
    {
        int ans=1;
        for(int i=0; i<n; i++)
        {
            cin>>x>>y;
            p[i].x1=x-sqrt(d*d-y*y);
            p[i].x2=x+sqrt(d*d-y*y);
            if(y>d||y<0||d<=0)
                ans=-1;
        }
        sort(p,p+n,cmp);
        double xx=p[0].x2;
        for(int i=1; i<n && ans!=-1; i++)
        {
            if(p[i].x1>xx)
            {
                ans++;
                xx=p[i].x2;
            }
            else if(p[i].x2<xx)
                {
                    xx=p[i].x2;
                }
        }
        printf("Case %d: %d\n",++kase,ans);
    }
}