POJ1328:Radar Installation

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Radar Installation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 41891   Accepted: 9270 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

=====================================题目大意=====================================

以海岸线为X轴（上方为海下方为岸）建立直角坐标系，给出海中N个岛屿（看做是点）的X-Y坐标。

求解至少需要在海岸线上安装多少个覆盖半径为R的雷达装置才能覆盖所有岛屿。

=====================================算法分析=====================================

贪心算法。

1、对每个岛屿：

   如果：岛屿纵坐标大于雷达覆盖半径，则不存在可以覆盖该岛屿的雷达放置位置（此时可直接得出答案为-1）。

   否则：以岛屿为圆心，雷达覆盖半径为半径做圆，交海岸线于(A,0)，(B,0)（A<=B）两点，则区间[A,B]为可以覆盖该岛屿的雷达

         放置区间。

2、题目转换为：

   选取尽可能少的点（即雷达放置位置）使得每个上述的雷达放置区间中都有点存在（从而每个岛屿都有雷达覆盖）。

   则使用贪心算法即可解决问题（此即刘汝佳白书P153：区间选点问题）。

=======================================代码=======================================

#include<cmath>
#include<cstdio>
#include<algorithm>

using namespace std;

int N,R;

struct Interval { double sta,end; } Inter[1005];

bool cmp(Interval& I1,Interval& I2)  
{
	return (I1.end<I2.end)||(I1.end==I2.end&&I1.sta>I2.sta);
}

bool ReadAndDealData()                  //读取并处理数据
{
	bool flag=true;                     //数据合理标志
	for(int i=0;i<N;++i)
	{
		double x,y;
		scanf("%lf%lf",&x,&y);
		if(flag==true)                  //只有在未发现不合理数据时才继续处理
		{
			if(R<y)                     //存在不合理数据
			{ 
				flag=false; 
			}
			else                        //计算可以覆盖第i个岛屿的雷达放置区间
			{
				Inter[i].sta=x-sqrt(R*R-y*y);
				Inter[i].end=2*x-Inter[i].sta;
			}
		}
	}
	return flag;
}

int main()
{
	for(int cas=1;scanf("%d%d",&N,&R)==2&&(N||R);++cas)
	{
		int cnt=-1;
		if(ReadAndDealData()==true)     //只有在未发现不合理数据时才继续处理
		{
			sort(Inter,Inter+N,cmp);
			cnt=1;
			double coverend=Inter[0].end;
			for(int i=1;i<N;++i)        //贪心算法
			{
				if(coverend<Inter[i].sta)
				{
					++cnt; 
					coverend=Inter[i].end;
				}
			}
		}
		printf("Case %d: %d\n",cas,cnt);
	}
	return 0;
}