poj 1328 Radar Installation

Radar Installation
题目地址

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 86095 Accepted: 19253

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.

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

#include <iostream>#include <algorithm>#include <stdlib.h>#include <math.h>using namespace std;struct point{    double left, right;}p[2010], temp;bool operator < (point a, point b){    return a.left < b.left;}int main(){    int n;    double r;    int kase = 0;    while (cin >> n >> r && (n || r))    {        bool flag = false;        for (int i = 0; i < n; i++)        {            double a, b;            cin >> a >> b;            if (fabs(b) > r)            {                flag = true;            }            else            {                p[i].left = a * 1.0 - sqrt(r * r - b * b);                p[i].right = a * 1.0 + sqrt(r * r - b * b);            }        }        cout << "Case " << ++kase << ": ";        if (flag)        {            cout << -1 << endl;        }        else        {            int countt = 1;            sort(p, p + n);            temp = p[0];            for (int i = 1; i < n; i++)            {                if (p[i].left > temp.right)                {                    countt++;                    temp = p[i];                }                else if (p[i].right < temp.right)                {                    temp = p[i];                }            }            cout << countt << endl;        }    }}