POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43793		Accepted: 9704

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

这个题据说是贪心题，但是网上的解题报告看来看去，觉得更像是区间覆盖，然后自己就用这样的思路写了一下，结果就AC了，具体就是先将平面上的问题化成数轴上的问题，将平面每个点坐标得到后先进行判断，如果纵坐标大于雷达半径，则输出-1结束，如果不是，运用勾股定理计算在X轴修建雷达可以辐射的区间，用结构体point的两个值left以及right存储区间的左值以及右值，按左值从小到大对结构体point进行排序，将第一个雷达放置在最左边第一个区间内，并且判定位置放到第一个区间的右值，判断下一个点的左值是否在判定点左面，如果在，则此雷达可以同时扫描到该点，继续判定右值是否在该判定点左面，如果在，将判定点变为该点的右值，如果左值不在判定点左面，则需添加雷达，雷达数量加1，并且将判定点变为该点右值，直到结束所有点的判定后输出当前雷达数量。

下面是AC的代码：

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
struct st
{
double left;
double right;
}point[1005];
int x[1005],y[1005];
void sort(int n)
{
   int i,j,k;
   st temp;
    for (i=0;i<n;i++)
        {
        k=i;
        for (j=i;j<n;j++) if (point[j].left<point[k].left) k=j;
        temp=point[i];point[i]=point[k];point[k]=temp;
        }
}
int main()
{
	int i,time=1,num,flag,n,d;
	double r,temp;
	while(1)
	{
		num=1;
		flag=1;
		scanf("%d%d",&n,&d);
		if(n==0&&d==0)
			break;
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&x[i],&y[i]);
			if(y[i]>d)
				flag=0;
		}
		if(flag==0)
		{
			printf("Case %d: -1\n",time);
			time++;
		}
		else
		{
			for(i=0;i<n;i++)
			{
				temp=sqrt(d*d-y[i]*y[i]);
			    point[i].left=(double)x[i]-temp;
			    point[i].right=(double)x[i]+temp;
			}
			sort(n);
			r=point[0].right;
			for(i=1;i<n;i++)
			{
				if(point[i].left<=r)
				{
					if(point[i].right<=r)
						r=point[i].right;
				}
				else
				{
					r=point[i].right;
					num++;
				}
			}
			printf("Case %d: %d\n",time,num);
			time++;
		}
	}
	return 0;
}