poj 1328 Radar Installation

题目链接：http://poj.org/problem?id=1328

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 67443		Accepted: 15121

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

Beijing 2002

 

题意： 输入n，d，表示有n个点，半径为d，接下来输入n行，表示n个点的坐标，最少需要几个才能将所有点包围。

 

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

using namespace std;

struct node
{
    double x,y;
} a[1005];

struct len
{
    double l, r;
} ans[1005];

bool cmp(len x1, len x2)
{
    if(fabs(x1.l-x2.l)<1e-10)
        return x1.r < x2.r;
    return x1.l < x2.l;
}

int main()
{
    int n;
    double d;
    int k=1;
    while(scanf("%d%lf",&n, &d),n+d)
    {
        int flag = 0;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf",&a[i].x,&a[i].y);
            if(a[i].y-d > 1e-10)
            {
                flag = 1;
            }
        }
        int s = 1;
        if(flag)
            s=-1;
        else
        {
            for(int i=0; i<n; i++)
            {
                double l = sqrt(pow(d,2) - pow(a[i].y,2));
                ans[i].l = a[i].x - l;
                ans[i].r = a[i].x + l;
            }
            sort(ans, ans+n, cmp);

            double R = ans[0].r;
            for(int i=1; i<n; i++)
            {
                if(ans[i].l-R > 1e-10)
                {
                    s++;
                    R = ans[i].r;
                }
                else
                {
                    R = min(R, ans[i].r);
                }
            }
        }
        printf("Case %d: %d\n",k++,s);
    }

    return 0;
}