Radar Installation（贪心）

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最新推荐文章于 2023-11-28 21:44:30 发布

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分类专栏：大菜鸟小历程贪心

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本文介绍了一个关于雷达安装的算法问题，旨在寻找覆盖所有岛屿所需的最小雷达数量。通过将问题抽象为一系列区间的覆盖问题，利用排序和贪心策略确定最佳雷达位置。

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Radar Installation

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

Submit Status Practice POJ 1328

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

题解：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目。其实转化为数学问题，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，都可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内至少存在一个点

代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
struct node
{
	double left;
	double right;
}a[1010];
bool cmp(node x,node y)
{
	return x.left<y.left;
}
int main()
{
	int n,k=0;
	double d;
	while(scanf("%d %lf",&n,&d)&&(n||d))
	{
		double x,y,cnt;
		int ans=1;
		bool flag=0;
		for(int i=0;i<n;i++)
		{
			scanf("%lf %lf",&x,&y);
			a[i].left=x-sqrt(d*d-y*y); //用sqrt的时候, 强制转换成double, 或者+0.0就可以
			a[i].right=x+sqrt(d*d-y*y);
			if(y>d||d<0)	flag=1;
		}
		if(flag)
			printf("Case %d: -1\n",++k);
		else
		{
			sort(a,a+n,cmp);   //sort里面的这个 n 必须是 int型的  表示指针的位移数
			cnt=a[0].right;
			for(int i=1;i<n;i++)
			{
				if(a[i].right<=cnt)
				{
					cnt=a[i].right;
				}
				else if(a[i].left>cnt)
				{
					cnt=a[i].right;
					ans++;
				}
			}
			printf("Case %d: %d\n",++k,ans);
		}
	}
	return 0;
}