Radar Installation POJ - 1328

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


题解：
　　直接考虑这个题目可能无从下手。
　　我们把模型抽象出来，就是从选取若干从以x轴为圆心，d为半径的圆，覆盖所有的所给的点，求最小的圆数。
　　这个模型无法下手，考虑转化模型，因为只能在x轴上，所以考虑对于每个点，求出，在x轴上所对应的可以覆盖这个点的区间段，那么问题就变成了对于每个区间，求最小点数可以覆盖所有的区间。经典贪心问题，剩下的就没什么了。
代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#define MAXN 10000
using namespace std;
struct qvjian{
    double l,r;
}a[MAXN];
int n;double d;

bool cmp(qvjian x,qvjian y){
    return x.r<y.r;
}

int main()
{
    int Case=0,flag=0;
    while(1){
        //memset(a,0,sizeof(a));
        cin>>n>>d;flag=0;
        if(!n&&!d) break;
        if(d<0) flag=1;
        for(int i=1;i<=n;i++){
            double x,y;
            scanf("%lf%lf",&x,&y);
            if(d*d-y*y<0){
                flag=1;
            }
            double l=x-sqrt(d*d-y*y),r=x+sqrt(d*d-y*y);
            a[i].l=l,a[i].r=r;
        }
        if(flag){
            printf("Case %d: -1\n",++Case);
            continue;
        }
        sort(a+1,a+n+1,cmp);
        int ans=0;double now=-(1<<30);
        for(int i=1;i<=n;i++){
            if(now<a[i].l){
                now=a[i].r;
            ans++;
            }
        }
        printf("Case %d: %d\n",++Case,ans);
    }
    return 0;
}