POJ-1328 Radar Installation-优快云博客

本文链接：https://blog.youkuaiyun.com/Poseidon__ming/article/details/52006513

本文介绍了一种通过计算几何的方法解决雷达安装问题的算法。该算法的目标是在海岸线上设置最少数量的雷达站，以覆盖海面上的所有岛屿。通过计算每个岛屿能够被雷达覆盖的最左和最右边界，并使用贪婪策略选择最佳位置，最终实现雷达站数量的最小化。

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Radar Installation
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 75484 Accepted: 16905

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

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

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

思路:
相对原点而言，我们假定x负半周方向为“左”，x正半轴方向为“右”。
我们找一个岛屿能被侦测到的极限范围，在雷达侦测区（圆）的左半圆上或者在雷达侦测区（圆）的右半圆上，换句话说当岛屿到雷达的距离等于d时，
雷达可以位于岛屿的左侧也可以位于雷达的右侧。而这就可以分别确定雷达相对与岛屿x的最左坐标和x最右坐标。
最左为：x - sqrt(d*d-y*y); 最右为：x - sqrt(d*d-y*y);
每个岛屿都有这样的最左和最右可被侦测坐标。
根据贪婪的思想，每次都应该将最右可被侦测坐标作为衡量标准。
假定当前的岛屿为p，当前的下一个为next。
1.如果next的最左可被侦测坐标比p的最右都大的话，只能再设一个雷达来侦测next了。
2.如果next的最左可被侦测坐标比p的最右小，这时会有两种情况。
A.next最右 < cur最右
B.next最右 >= cur最右
对于B情况,我们可以直接侦测到next了, 可以找next的next了.
对于A情况,也就等价于next包含于p, 这样就应该把next的右最为衡量标准了.
因为这样可以左移最右坐标, 可以让可能更多的岛屿被侦测到(他们的最左与衡量标准有更多的交集)

提供一组测试数据：
20 8
-20 7
-18 6
-5 8
-21 8
-15 7
-17 5
-1 5
-2 3
-9 6
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 7
9 6
10 5
0 0

答案是 4.

代码如下：

#include <cstdio>
#include <algorithm>
#include<math.h>
#include<string.h>
using namespace std;
struct dao
{
    double x,y,l,r;
}p[1005];
bool cmp(dao a,dao b)
{
    return a.l<b.l;
}
int main()
{
    int n,d,a=1;
    while(~scanf("%d%d",&n,&d)&&n||d)
    {
        int sum=1;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
            p[i].l=p[i].x-sqrt(fabs(d*d-p[i].y*p[i].y));
            p[i].r=p[i].x+sqrt(fabs(d*d-p[i].y*p[i].y));
            if(d<p[i].y) sum=0;
        }
        if(!sum)
        {
            printf("Case %d: -1\n",a++);
            continue;
        }
        sort(p,p+n,cmp);
        double z=p[0].r;
        for(int i=1;i<n;i++)
        {
            if(z<p[i].l)
            {
                sum++;
                z=p[i].r;
            }
            else
            {
                if(p[i].r<z) z=p[i].r;
            }
        }
        printf("Case %d: %d\n",a++,sum);
    }
    return 0;
}