Radar Installation poj1328

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解决在二维平面上，如何使用最少数量的雷达装置覆盖所有岛屿的问题。通过将问题转化为区间覆盖，利用贪心算法确定最佳雷达位置。

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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.

在这里插入图片描述

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

不得不承认北大的题目质量是真的高。这个问题用到了贪心。刚开始。我考虑的是找到最大的y 当然这个y是小于系统的半径的。以这个最大的y为半径画圆。然后找出这个范围内的点。并删除。
后来参考了别人的想法以后便有了新的想法。
我们可以以一个点。做圆心以系统的检测半径为半径画圆。这样便会跟坐标轴出现两个交点。这样我们只要在这两个交点的范围内随便取一个点就可以检测这个点。当我们把所有点与坐标轴的交点求出来。题目便成了一个类似区间覆盖的问题了。只需要求是否有公共点了，问题就迎刃而解了。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
struct node
{
    double x,y;
} da[10000];
bool cmp(node a,node b)//排序
{
    if(a.y==b.y)
        return a.x>b.x;
    return a.y<b.y;
}
int main()
{
    int n,d;
    double x1,y1;
    int ans;
    int k=0;
    while(scanf("%d%d",&n,&d)&&n||d)
    {
        ans=1;
        for(int i=0; i<n; i++)//注意是double
        {
            scanf("%lf%lf",&x1,&y1);
            da[i].x=x1-sqrt(d*d-y1*y1);
            da[i].y=x1+sqrt(d*d-y1*y1);
            if((d-y1)<0||d<0)
                ans=-1;
        }
        if(ans==-1)
        {
            printf("Case %d: -1\n",++k);
            continue;
        }
        else
        {
            double t1;
            sort(da,da+n,cmp);
            t1= da[0].y;
            for(int i=1; i<n ; i++)
                if(da[i].x>t1)
                {
                    ans++;
                    t1 = da[i].y;//更新右范围
                }

            printf("Case %d: %d\n",++k, ans);
        }
    }
}