D - Radar Installation POJ - 1328

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本文介绍了一个关于使用雷达装置覆盖海上岛屿的问题。通过建立坐标系，确定雷达覆盖范围，并利用贪心算法来解决如何用最少数量的雷达装置实现全部岛屿的有效覆盖。

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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.

这里写图片描述

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

题意：
岛屿在上面，陆地在下面，安装雷达覆盖岛屿，求需要最少的雷达。

思路：
求每一个岛屿的覆盖面积，然后减去重复的。就是贪心。

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
struct stuff
{
    double x,y;
};
bool cmp(stuff a,stuff b)
{
    return a.x<b.x;
}
int main()
{
    int n,d,q=1;
    stuff a[1005];
    while(cin>>n>>d)
    {
        if(n==0&&d==0)
            break;
        int flag=0;
        int x,y;
        for(int i=0;i<n;i++)
        {
            cin>>x>>y;
            if(y>d) flag=1;
            a[i].x=(double)x-sqrt((double)d*d-y*y);
            a[i].y=(double)x+sqrt((double)d*d-y*y);
        }
        cout<<"Case "<<q++<<": ";
        if(flag)
        {
            cout<<"-1"<<endl;
            continue;
        }
        sort(a,a+n,cmp);
        int sum=1;
        double now=a[0].y;
        for(int i=1;i<n;i++)
        {
            if(a[i].x>now)
            {
                sum++;
                now=a[i].y;
            }
            else if(a[i].y<=now)
            {
                now=a[i].y;
            }
        }
        cout<<sum<<endl;
    }
}