POJ1328 Radar Installation 贪心

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 69114		Accepted: 15506

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

题目大意:在平面上有n个点,告诉你圆的半径为d,求在x轴上最少需要几个点可以使得以这些点为圆心,半径为d的圆覆盖所有的n个点

思路：对于每一个点，找出他所对应的最左雷达位置与最右雷达位置，便有n个区间,从左向右遍历判断下一个区间能否与上一个区间的交集重合，若能重合则可共用同一个雷达。若能公用一个雷达则更新交集范围。注意d<0的情形，另外若|y|>d则可判断-1

实现代码：

//************************************************************************//
//*Author : Handsome How		                                 *//
//************************************************************************//
//#pragma comment(linker, "/STA	CK:1024000000,1024000000")
#pragma warning(disable:4996) 
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <complex>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>                                                
#include <cassert>
#define fur(i,a,b) for(int i=(a);i<=(b);i++)
#define furr(i,a,b) for(int i=(a);i>=(b);i--)
using namespace std;
typedef long long LL;
const int MAXN = 1000 + 20;
struct Node
{
	double l;
	double r;
}len[MAXN];
bool compare(Node A, Node B)
{
	if (A.l != B.l)return A.l < B.l;
	else return A.r < B.r;
}
int main()
{
	//freopen("E:\\data.in", "r", stdin);
	//freopen("E:\\data.out", "w", stdout);
	int n,kase=0;
	double d;
	while (scanf("%d%lfd", &n,&d) != EOF)
	{
		if(n == 0 && d == 0)break;
		printf("Case %d: ",++kase);
		bool legal = true;
		double x, y,t;
		fur(i, 0, n-1) 
		{
			scanf("%lf%lf", &x,&y);
			if (y > d || y < d*(-1.0))legal = false;		 //最高点能否被雷达范围覆盖
			t = sqrt(d*d - y*y);
			len[i].l = x - t;
			len[i].r = x + t;
		}
		if (d < 0) { printf("-1\n"); continue; }
		 if(legal==false) { printf("-1\n"); continue; }
		 sort(len, len + n, compare);
		 //fur(i, 0, n - 1)printf("%lf %lf\n", len[i].l, len[i].r);
		 int last = 0,cnt=1;
		 fur(i, 1, n - 1)
		 {
			 if (len[i].l <= len[last].r) 
			 { 
				 len[last].l = max(len[last].l, len[i].l); 
				 len[last].r = min(len[last].r, len[i].r);
					 continue;							   //交集的更新
			 }
			 last = i;									   //新开一个区间
			 cnt++;										  
		 }
		 printf("%d\n",cnt);
	}
	return 0;
}