poj 1328Radar Installation

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


这题的大意就是给你land的坐标，现在要建立雷达站，问怎样建立使得用最少的雷达包含所有的land
这题是第一个专题想的最久的一题= =训练计划上写的是贪心思想。可就是没想到怎么个操作。。
后来在纸上画画，才发现了突破点。以每个站点为圆心，d为半径画圆，会与x轴相交，若相离，正好说明了不存在的情况。通过相交的点的坐标来确定该点周围
建立雷达的最左和最右范围。每两个半圆会出现相离，相交和内嵌的情况，前两者可以以最小的那个lft的rgt为出发点，只要包含于它左边的不用记，因为肯
定能涉及到，如果一个lft大于了这个rgt，说明有新的出现，更新即可=。=对于第三种情况，如果又一半圆与内嵌的圆相交，如果按照前面方法做的话，会
少建立一个。所以此时必须改变它的rgt，使他是内嵌的那个小圆的最右值。。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
struct island
{
    double lft;
    double rgt;
}p[1100];
bool cmp(island a,island b)
{
    return a.lft==b.lft?a.rgt<b.rgt:a.lft<b.lft;
}
int main()
{
    int n,i,cas=1;
    double d,x,y,z;
    while(scanf("%d%lf",&n,&d)&&(d+n))
    {
        int flag=1;
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf",&x,&y);
            if(y>d)
            {
                flag=0;
                continue;
            }
            z=d*d-y*y;
            p[i].lft=x-sqrt(z);
            p[i].rgt=x+sqrt(z);
        }
        printf("Case %d: ",cas++);
        if(!flag)
        {
            printf("-1\n");
            continue;
        }
        sort(p,p+n,cmp);
        int cnt=0;
        double maxx=p[0].rgt;
        for(i=0;i<n;i++)
        {
            if(p[i].lft>maxx)
            {
                cnt++;
                maxx=p[i].rgt;
            }
            if(p[i].rgt<maxx)
                maxx=p[i].rgt;
        }
        printf("%d\n",cnt+1);
    }
    return 0;
}