poj 1328 Radar Installation

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

求每个岛对应的覆盖它的雷达的有效区间，然后扫描区间去掉重合的。

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
struct Island
{
	int x; // 岛屿x坐标
	int y; // 岛屿y坐标
	double minRadarX; // 覆盖它的雷达有效区间的左端
	double maxRadarX; // 覆盖它的雷达有效区间的右端
};

bool IslandCmp(Island & island1, Island & island2)
{
	return island1.x < island2.x;
}
int main()
{
	int n, d;
	int caseNo = 0;
	vector<Island> islandsVec;
	while(cin >> n >> d)
	{
		islandsVec.clear();
		if(n == 0 || d == 0) 
			break;
		caseNo++;
		Island island;
		for(int i = 0; i < n; i++)
		{
			cin >> island.x >> island.y;
			islandsVec.push_back(island);
		}

		// 按岛屿x坐标排序
		sort(islandsVec.begin(), islandsVec.end(), IslandCmp);

		// 求每个岛屿对应的雷达的有效区间
		bool bNoAnswer = false;
		vector<Island>::iterator iter;
		for(iter = islandsVec.begin(); iter != islandsVec.end(); iter++)
		{
			if(iter->y > d)
			{
				bNoAnswer = true;
				break;
			}
			double r = sqrt(d * d * 1.0 - iter->y * iter->y);
			iter->minRadarX = iter->x - r;
			iter->maxRadarX = iter->x + r;
		}

		if(bNoAnswer)
		{
			cout << "Case " << caseNo << ": " << -1 << endl;
		}
		else
		{
			// 统计实际需要的雷达数目
			int result = 0;
			double preEnd;
			vector<Island>::iterator iter;
			for(iter = islandsVec.begin(); iter != islandsVec.end(); iter++)
			{
				if(iter == islandsVec.begin())
				{
					result++;
					preEnd = iter->maxRadarX;
				}
				else
				{
					if(iter->minRadarX > preEnd)
					{
						result++;
						preEnd = iter->maxRadarX;
					}
					else
					{
						if(iter->maxRadarX < preEnd)
							preEnd = iter->maxRadarX;
					}
				}
			}
			cout << "Case " << caseNo << ": " << result << endl;
		}
	}
	return 0;
}