Radar Installation

Radar Installation

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 93   Accepted Submission(s) : 46

Problem 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

 

题目大意：

先输入两个数字n与d，n代表小岛的个数，d代表卫星所能覆盖范围的半径，接下来n行数字，代表n个小岛的坐标，卫星安置在x轴上，求最少需要几个卫星，能够覆盖所有的小岛

 

思路：

以小岛为中心，以卫星半径为半径画圆，该圆与x轴所交的弦即为可放置卫星的地方，若想放置卫星最少，则应放置在右边界地方，请仔细思考

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
struct qj
{
    double l, r;
}b[1110];//记录可以放置卫星的地方的左右区间
bool cmp(qj a, qj b)
{
    if(a.r == b.r)
        return a.l > b.l;
    return a.r < b.r;
}//自定义排序，按照右区间从小到大排列，若右区间相等，则按左区间从大到小排列
int main()
{
    double x, y;
    double d,temp;
    int  i, n, count, num=1;
    while(~scanf("%d %lf", &n, &d), n&&d)
    {
        for(i = 0; i < n; i ++)
        {
            scanf("%lf %lf", &x, &y);
            b[i].l = x - sqrt(d*d - y*y);//放置卫星的地方的左区间
            b[i].r = x + sqrt(d*d - y*y);//放置卫星地方的右区间
        }
        sort(b, b + n, cmp);
        count = 1;
        temp = b[0].r;
        for(i = 1; i < n; i ++)
        {
            if(b[i].l > temp)//判断两个区间是否有交叉
            {
                temp = b[i].r;
                count++;
            } 
        }
        printf("Case %d: %d
", num++, count);
    }
    return 0;
}