Radar Installation

描述

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

输入

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

输出

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.

样例输入

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3 2
1 2
-3 1
2 1

1 2
0 2

0 0

样例输出

1
2
Case 1: 2
Case 2: 1

代码

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#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <vector>

using namespace std;

int d, n;

struct point {
int x, y;
double start, end;

// Here we assume y is always smaller than d
// We'll check whether y is larger than d when input
point(int x, int y) : x(x), y(y)
{
double range = sqrt(pow(double(d), 2.0) - pow(double(y), 2.0));
start = double(x) - range;
end = double(x) + range;
}

bool operator<(const point &p) const
{
return start < p.start;
}

bool operator>(const point &p) const
{
return start > p.start;
}
};

int main(void)
{
cin >> n >> d;
int case_num = 0;
while (n != 0 && d != 0) {
case_num++;
vector<point> v;
int x, y;
bool has_result = true;
for (int i = 0; i < n; i++) {
cin >> x >> y;
if (y > d) {
has_result = false;
}
v.push_back(point(x, y));
}
if (!has_result) {
cout << "Case " << case_num << ": " << -1 << endl;
}
else {
sort(v.begin(), v.end());
double radar_x = -1000000000;
int radar_num = 0;
for (vector<point>::iterator it = v.begin(); it != v.end(); it++) {
if (it->start > radar_x) {
radar_num++;
radar_x = it->end;
}
else {
if (it->end < radar_x) {
radar_x = it->end;
}
}
}
cout << "Case " << case_num << ": " << radar_num << endl;
}
cin >> n >> d;
}
return 0;
}

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