poj 1328 Radar Installation 贪心区间选点问题

Posted 2020-06-23 gcczhongduan

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54798		Accepted: 12352

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

题意：就是找最少的站，来覆盖所有的点。

思路：我们能够以点来做半径为d的圆，与x轴的相交，假设不相交那么肯定完不成任务，反之就转化成了区间选点问题。

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define M 1005


struct node {
	double st, en;
}s[M];

int cmp(node a, node b){
	if(a.en == b.en) return a.st > b.st;
	return a.en<b.en;	
}

int main(){
	int n, v = 1; double d;
	while(scanf("%d%lf", &n, &d), n||d){
		int i, j;
		double a, b;
		int flag = 0;
		for(i = 0; i < n; i ++){
			scanf("%lf%lf", &a, &b);
			if(b>d) flag = 1;
			if(flag == 0){
				s[i].en = a+sqrt(d*d-b*b);
				s[i].st = a-sqrt(d*d-b*b);
				//printf("%lf %lf %d..\n", s[i].st, s[i].en, i);
			}
		//	scanf("%lf%lf", &s[i].st, &s[i].en);
		}
		printf("Case %d: ", v++);
		if(flag){
			printf("-1\n"); continue;
		}
		sort(s, s+n, cmp);
		int ans = 1;
		double maxr = s[0].en;
		i = 1, j = 0;
		while(i < n){
			if(s[i].st <= s[j].en){
			//if(maxr < s[i].en) maxr = s[i].en;
				++i;
			}
			else {
				//if(j == i-1) 
				j = i;
				++ans;
			//	maxr = s[i].en;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}