NYOJ 287 Radar

Radar

时间限制：1000 ms  |           内存限制：65535 KB

难度：3

 

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.

 

 

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

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

样例输出

Case 1: 2
Case 2: 1
思路：确定岛在x轴上可覆盖区间（即在此范围内，信号可覆盖岛）右边界排序；
代码：

 
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int MAXN=1005;
struct Line
{
    double l,r;
}line[MAXN];//每个岛作半径为d的圆，截得区间
bool cmp(Line a,Line b)
{
    return a.r<b.r;
}
int main()
{
    int n,d;
    int i;
    int x,y;
    int num=1,flag,count;
    while(scanf("%d %d",&n,&d)&&n&&d)
    {
        flag=1;
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&x,&y);
            if(!flag) continue;
            if(y<=d){
              line[i].l=(double)x-sqrt((double)d*d-y*y);
               line[i].r=(double)x+sqrt((double)d*d-y*y);
            }else flag=0;
        }
        if(!flag)
        {
            printf("Case %d: -1
",num++);
            continue;
        }
        sort(line,line+n,cmp);
        count=1;
        double now=line[0].r;
        for(i=1;i<n;i++)
        {

            if(line[i].l<=now+0.00005) continue;
            now=line[i].r;
            count++;
        }
      printf("Case %d: %d
",num++,count);
    }
    return 0;
}