POJ1328 Radar Installation（贪心）

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

Source

Beijing 2002

题目链接：poj1328

         首先这个题的题意就是在一个平面坐标内，有n个处于x轴上方的点，作为岛屿。在X轴上可以安放雷达，雷达的半径为d。题目要求根据n个岛屿的坐标和雷达的半径d求解出最少需要多少个雷达才嫩把所有的岛屿包括进来。如果没有办法把所有的包括进来，则出输出-1.

         第一次看到这个题我的想法和后来在网上找的差不多，我想的是将每个坐标（x0,y0）与X轴的交点求出来，求法是勾股定理x=x0-根号下（d²-y0²），然后画出一个区间[xd,yd]，比较这n个岛屿的区间之间是否存在重叠，若存在则安装一个雷达。

        但是后来想了想可行度不高，于是看了下网上的解析，其实就是一个数据结构设置的不当的问题，网上是将每个点与X轴的左右交点存储在一个结构体数组中，然后排序，比较左右两岛屿的左右两个交点的左右关系。并用temp来标记稍微靠右边的那个点，用p[i]来更新新的点，直到temp.right<p[i].left为止，再让temp=p[i]，进行新的比较。还是这种方法比较好实现。

下面是代码和一部分解析。（转载）

图中ABCD为海岛的位置。假设本题半径为2（符合坐标系），那么A点坐标为（1，1）以此类推。

在题中，记录每个点的坐标，并把一个点新加一个标记变量以标记是否被访问过。

首先以A为圆心，r为半径做圆，交x轴与E1（右）点，做出如图中绿色虚线圆。然后以E1为圆心，半径为r做圆。看此时下一点（B）是否在该圆中。如果在，那么将该点标记变量变为true，再判下一点（C），如果不在那么就新增一个雷达。

后来，换了思路，存储图中红色圆与X轴的交点。

还是原来的贪心思路，仍然先排序，但排序的准则是按红色圆与x轴左交点的先后顺序。

如图，如果B点圆（且如此称呼）的左交点（J1点）在A点圆右交点（E1）左侧，那么B点一定涵盖在A点圆内部。再如图，如果D点圆的左交点（图中未标示）在A点圆的右交点（未标示）右，则D点不在A点圆中。

故，有如下代码：

#include <iostream>  
#include <algorithm>  
#include <stdlib.h>  
#include <math.h>  
  
using namespace std;  
  
struct point  
{  
    double left, right;  
}p[2010], temp;  
  
bool operator < (point a, point b)  
{  
    return a.left < b.left;  
}  
  
int main()  
{  
    int n;  
    double r;  
    int kase = 0;  
    while (cin >> n >> r && (n || r))  
    {  
        bool flag = false;  
        for (int i = 0; i < n; i++)  
        {  
            double a, b;  
            cin >> a >> b;  
            if (fabs(b) > r)  
            {  
                flag = true;  
            }  
            else  
            {  
                p[i].left = a * 1.0 - sqrt(r * r - b * b);  
                p[i].right = a * 1.0 + sqrt(r * r - b * b);  
            }  
        }  
        cout << "Case " << ++kase << ": ";  
        if (flag)  
        {  
            cout << -1 << endl;  
        }  
        else  
        {  
            int countt = 1;  
            sort(p, p + n);  
            temp = p[0];  
              
            for (int i = 1; i < n; i++)  
            {  
                if (p[i].left > temp.right)  
                {  
                    countt++;  
                    temp = p[i];  
                }  
                else if (p[i].right < temp.right)  
                {  
                    temp = p[i];  
                }  
            }  
            cout << countt << endl;  
        }  
    }  
}