Radar Installation---（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 115873		Accepted: 25574

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

分析：首先根据小岛的坐标计算出每座小岛对应海岸线上的范围。将每个小岛对应在海岸线上的范围进行排序，使得每个雷达范围的最小值进行递增。

对雷达范围进行贪心。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=1010;
#define INf 0x3f3f3f3f
struct node{
    double l,r;
}point[maxn];

bool cmp(const node &a,const node &b){
    return a.l<b.l;
}

int main(){
    int n,d;
    int case1=0;
    while(~scanf("%d%d",&n,&d)&&n){
        int flag=0;
        for(int i=0; i<n; i++ ){
            int x,y;
            cin>>x>>y;
            if(y>d){
                flag=1;
//                break;
            }
            double p=sqrt((double)(d*d)-y*y);
            point[i].l=x-p;
            point[i].r=x+p;
        }
        printf("Case %d: ",++case1);
        if(flag){
            cout<<-1<<endl;
            continue;
        }
        sort(point,point+n,cmp);
        int ans=1;
        node tmp=point[0];
        for( int i=1; i<n; i++ ){
            if(tmp.r>=point[i].r) tmp=point[i];
            else if(tmp.r<point[i].l){
                ans++;
                tmp=point[i];
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}