Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 135401 | Accepted: 29732 |
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
Solution
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
typedef pair<double, double> p;
int n, d, k = 0;
while (scanf("%d%d", &n, &d) == 2 && (n || d))
{
vector<p> intv;
bool flag = false;
for (int i = 0; i < n; i++)
{
int x, y;
scanf("%d%d", &x, &y);
if (d < y) flag = true;
double x1 = x - sqrt(d * d - y * y);
double x2 = x + sqrt(d * d - y * y);
intv.push_back(p(x1, x2));
}
if (flag) printf("Case %d: %d\n", ++k, -1);
else
{
sort(intv.begin(), intv.end());
int cnt = 1;
double r_last = intv[0].second;
for (int i = 1; i < n; i++)
{
double l_now = intv[i].first;
double r_now = intv[i].second;
if (r_last >= l_now) r_last = min(r_last, r_now);
else
{
cnt++;
r_last = r_now;
}
}
printf("Case %d: %d\n", ++k, cnt);
}
}
return 0;
}