Radar Installation(贪心，可以转化为今年暑假不ac类型)

Radar Installation

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

Total Submission(s) : 54   Accepted Submission(s) : 28

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

题解：先转化为区间点，再排序，区间找点；

代码：

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct Node{
    double s,e;
};
int n,d,k;
Node area[1010];
int cmp(Node a,Node b){
    return a.e<b.e;
}
int change(int x,int y){
    if(y>d)return 0;
    double a,b,m=sqrt(d*d-y*y);
    a=x-m;b=x+m;
    area[k].s=a;area[k].e=b;k++;
    return 1;
}
int main(){int t,x,y,flot,temp,num,l=0;
    while(scanf("%d%d",&n,&d),n||d){k=0;flot=1;temp=0;num=1;l++;
            for(int i=0;i<n;i++){
                scanf("%d%d",&x,&y);
                t=change(x,y);
                if(!t)flot=0;
            }
            sort(area,area+k,cmp);
            for(int i=0;i<k;i++){
                if(area[i].s>area[temp].e)temp=i,num++;
            }
            printf("Case %d: %d\n",l,num);
    }
    return 0;
}