贪心 1011

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. <br> <br>The input is terminated by a line containing pair of zeros <br>

 

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
雷达
区间问题
struct 设立区间
然后贪心比较最后
然后排列前面
逐个缩小
直到缩小没有
找下一个
#include <cstdio>
#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<math.h>
#include<string.h>
#include<map>
#include<set>
#include<vector>
#include<iomanip>
using namespace std;
struct ww
{
    double mix;
    double max;
};
    bool cmp(const ww &A,const ww &B)
    {
     if(A.max<=B.max) return true;
     return false;
    }

int main()
{   double paopao;
    int n;
    int t;
    int s[1000];
    int aa=0;
    ww l[1000];
    while(cin>>n>>t&&t!=0||n!=0)
    {    aa=aa+1;
        double a;
        double b;
        memset(s,0,sizeof(s));

        int i;
        int flag=0;
        for(i=1;i<=n;i++)
        {
            cin>>a;
            cin>>b;
            paopao=sqrt(t*t-b*b);
            l[i].mix=a-paopao;
            l[i].max=a+paopao;
            if(b>t)
            flag=1;
        }

        if(flag==1)
        cout<<"Case "<<aa<<": "<<"-1"<<endl;
        else
        {
       sort(l+1,l+n+1,cmp);
       int xx=n;
       int sum=1;
       for(i=n-1;i>=1;i--)
       {
           if(l[i].max<l[xx].mix)
           {
               sum++;
               xx=i;
           }
           else if(l[i].mix>=l[xx].mix)
            {  xx=i;
            }
       }

       cout<<"Case "<<aa<<": "<<sum<<endl;
        }
    }
    return 0;
}