POJ 1328 Radar Installation

雷达覆盖问题，贪心算法。

题目来源：CSUST2012年8月暑假集训组队后第一场个人赛（貌似是8月7日）

原题链接：http://poj.org/problem?id=1328

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 34841		Accepted: 7737

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

 

/******************************************
题意：给你一个直角坐标系;
      在x轴上方给你n个小岛，再给出雷达的覆盖半径d;
      当n和d为0时测试结束;
      剩下n行，依次给出岛屿的坐标（测试数据是int型，注意double转换）;
      求在x轴上最少设置几个雷达，可以覆盖所有的岛屿，如果不能完全覆盖则输出结果为-1;
思路：1、已各个小岛小岛为圆心，以雷达覆盖半径为半径做园，记录与x轴的交点坐标（如果有交点则记录，如果有一个没有交点则表示不能完全覆盖）;
      2、根据各个岛屿与x轴的左右交点排序（先按照左交点从小到大排序，如果坐交点相同，则按照又交点从小到大排序）;
      3、那么经过上面的步骤后，各个交点区肯定会有相加范围，尽量在相加最多的范围设置雷达即可;
      4、从左到右开始查找，设置一个点point为当前雷达覆盖指向的最右点，初始化为第一个圆的右交点;
      5、如果下一个圆的重叠区的左交点大于当前point，雷达数目+1，并且把当前的圆的右交点赋给point;
      6、如果下一个圆的重叠区的右交点小于当前point，说明它已经被覆盖，直接把它的右交点赋给point即可。
注意：由于要用到sqrt，所以要小心double型的转换。
***********************************************/
//Accepted	200K	16MS	C++	928B	2012-09-18 19:43:01
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct Node
{
	double left,right;
}node[1010];//圆与x轴的交点
int cmp(Node a,Node b)
{//排序：先按照与x轴的左交点从小到大排，如果左交点相同再按照右交点从小到大排
	return a.left<b.left || (a.left==b.left && a.right<=b.right);
}
int main()
{
	int n;
	int i;
	double d;
	double x,y;
	int sum,ok,c=0;
	double point;//记录当前雷达在x轴上覆盖区域的最右点
	while(scanf("%d%lf",&n,&d)!=EOF && n)
	{
		ok=1;
		if(d<0)//如果雷达覆盖半径<0，则不可能
			ok=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf%lf",&x,&y);
			if(y>d)//排除不可能情况
				ok=0;
			double tt=sqrt(d*d-y*y);//注意double的转换
			node[i].left=1.0*x-tt;//记录左交点
			node[i].right=1.0*x+tt;//记录右交点
		}
		if(ok==0)//如果不可能
		{
			printf("Case %d: -1\n",++c);
		}
		else
		{
			sort(node,node+n,cmp);
			point=node[0].right;
			sum=1;//设置第一个雷达
			for(i=1;i<n;i++)//挨个岛屿查找
			{
				if(node[i].left>point)//如果当前查找的岛屿的最左边的覆盖范围都不在雷达的覆盖范围内
				{
					sum++;//建立新雷达
					point=node[i].right;//以当前的岛屿的右交点赋给新建立的雷达的最右边的覆盖范围
				}
				else if(node[i].right<point)//当前岛屿的右交点都小于前面雷达覆盖位置，则说明已经被覆盖
					point=node[i].right;//更新当前雷达覆盖范围
			}
			printf("Case %d: %d\n",++c,sum);
		}
	}
	return 0;
}