POJ - 1328 Radar Installation 贪心-白红宇的个人博客

发布日期：2021-05-09 01:46:28 浏览次数：17 分类：博客文章

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Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 96177		Accepted: 21378

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

Source

可将点的坐标转为在x轴上对应的位置，然后根据右端对应的值进行排序判断。

#include
   
    #include
    
     #include
     
      using namespace std;struct pos{	double left,right;};bool flag;bool compare(pos a, pos b){	return a.right < b.right;}int main(){	int n;	double d;	int m = 1;	pos a[1005];	while (~scanf("%d %lf", &n, &d))	{		if (!n || !d)		{			break;		}		int sum = 1;		flag = true;		double x, y;		for (int i = 0; i
      
        d)			{				flag = false;				continue;			}			a[i].left = x - sqrt(d*d - y*y);			a[i].right = x + sqrt(d*d - y*y);		}				if (!flag)		{			printf("Case %d: -1\n",m);			m++;			continue;		}		sort(a, a + n,compare);		double temp = a[0].right;		for (int i = 1; i < n; i++)		{			if (temp

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