Radar Installation POJ-1328 解题报告

题目:POJ-1328

http://poj.org/problem?id=1328

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 102221		Accepted: 22734

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

解题分析

题目来源：POJ很好很有层次感

核心算法：贪心算法

  题目的解法是将每个岛屿需要的雷达安装范围抽象成一个线条，把所有线条按照左端点的先后顺序排序，然后再观察每个左端点：

如果第i个左端点可以覆盖之前所有未覆盖的岛屿，那么观察下一个；

如果第i+1个左端点开始不能覆盖之前的所有未覆盖断电，那么就在i处安装一个雷达

如果i+1可以覆盖，那么观察i+2，以此类推

值得注意的是边界值，y==d的情况和y==0的情况都是需要考虑的

（没错y可以等于0，我因为这个调了2个小时）

Ac码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <stack>
#include <cstring>
#include <math.h>
using namespace std;

int n,d;
struct Island {
	double x,y;
	double areal,arear;
	bool operator < (Island & land) const {
		return areal<land.areal;
	}
}island[1010]; 

bool Judge (int first,int last,double pos)  {
	for (int i=first;i<=last;++i)
		if(pos<island[i].areal||pos>island[i].arear)	return false;
	return true;
} 

int main()
{
	 int time=0;
	 while (cin>>n>>d) {
	 	time++;
	 	bool flag=0,flagone=0;
	 	int firstNoCover=0;
	 	int rader=0;
	 	if (n==0&&d==0)	break;
	 	if (n==1)	flagone=1;
	 	
	 	for(int i=0;i<n;++i) {
	 		cin>>island[i].x>>island[i].y;
	 		if(island[i].y>d)	flag=1;
			if(island[i].y<0)	flag=1;
			if(island[i].y==d){
				island[i].areal=island[i].x;
	 			island[i].arear=island[i].x;
			}
			else {
	 			island[i].areal=island[i].x-sqrt(d*d-island[i].y*island[i].y);
	 			island[i].arear=island[i].x+sqrt(d*d-island[i].y*island[i].y);
			}
		}
		if (flag)  {
		 	cout<<"Case "<<time<<": -1"<<endl;
		 	continue;
		}
		if(flagone)  {
			cout<<"Case "<<time<<": 1"<<endl;
			continue;
		}
		 
		sort(island,island+n);
		
		
		for (int i=1;i<n;++i)  {
			if(i==n-1)  {
				if(Judge(firstNoCover,i-1,island[i].areal))	
				{
					rader++;
				}
				else 
				{
					rader+=2;
				}
			}
			else {
				if(Judge(firstNoCover,i-1,island[i].areal))	continue;
				else {
					firstNoCover=i;
					rader++;
				}
			}
		}
	 	cout<<"Case "<<time<<": ";
		cout<<rader<<endl;
	 }
	 return 0;
}