POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 91342	Accepted: 20438

Description

Assume thecoasting is an infinite straight line. Land is in one side of coasting, sea inthe other. Each small island is a point locating in the sea side. And any radarinstallation, locating on the coasting, can only cover d distance, so an islandin the sea can be covered by a radius installation, if the distance betweenthem is at most d. 

We use Cartesian coordinate system, defining the coasting is the x-axis. Thesea side is above x-axis, and the land side below. Given the position of eachisland in the sea, and given the distance of the coverage of the radarinstallation, your task is to write a program to find the minimal number ofradar installations to cover all the islands. Note that the position of anisland is represented by its x-y coordinates. 

 
Figure A Sample Input of Radar Installations

 

Input

The input consistsof 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 isthe distance of coverage of the radar installation. This is followed by n lineseach containing two integers representing the coordinate of the position ofeach 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 caseoutput one line consisting of the test case number followed by the minimalnumber of radar installations needed. "-1" installation means nosolution 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

 

 

想法：貪心

統計所有島嶼在在x軸上可放置的範圍

class island {
public:
	int x, y;
	double l, r;
};

isl[i].l = isl[i].x*1.0 - sqrt(d*d - isl[i].y*isl[i].y);
isl[i].r = isl[i].x*1.0 + sqrt(d*d - isl[i].y*isl[i].y);

 

然後更加island.r的大小排序，先設置一個radar在island[0].r處，然後循環比較radar和island[i]更新

for (int i = 0; i<n; i++) {
			if (isl[i].r == radar)
				continue;
			else if (isl[i].l > radar) {
				radarNum++;
				radar = isl[i].r;
			}
		}

完整代碼：

//POJ 1328 Radar Installation

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define max 1000
using namespace std;
class island {
public:
	int x, y;
	double l, r;
};
bool cmp(island a, island b) {
	if (a.r == b.r) return a.l<b.l;
	return a.r<b.r;
}

int main(int argc, char const *argv[])
{
	freopen("in.txt", "r", stdin);
	int n, d, c = 1;
	island isl[max];
	while (cin >> n >> d &&(n||d)) {
		int radarNum = 1;
		bool flag = false;
		for (int i = 0; i<n; i++) {
			cin >> isl[i].x >> isl[i].y;
			if (isl[i].y > d) { flag = true; continue; }
			isl[i].l = isl[i].x*1.0 - sqrt(d*d - isl[i].y*isl[i].y);
			isl[i].r = isl[i].x*1.0 + sqrt(d*d - isl[i].y*isl[i].y);
		}
		if (flag) { printf("Case %d: -1\n", c++); continue; }
		sort(isl, isl + n, cmp);
		double radar = isl[0].r;
		for (int i = 0; i<n; i++) {
			if (isl[i].r == radar)
				continue;
			else if (isl[i].l > radar) {
				radarNum++;
				radar = isl[i].r;
			}
		}
		printf("Case %d: %d\n", c++, radarNum);
	}
	return 0;
}