1011 of greedy strategy

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. <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 21 2-3 12 11 20 20 0

 

Sample Output

Case 1: 2Case 2: 1

 

 题目要求：提供雷达半径和小岛位置坐标，计算最少的数目的雷达覆盖所有的小岛，若不能覆盖则输出-1.

题目思路：1进行输入的时候进行一次判断，若纵坐标大于雷达半径则标记。若已经标记则直接输出-1，否者算出每个小岛在x轴那些范围的雷达可以检测，把每组的开始坐标进行排序，然后从最小进行求最大重叠区间，当没有重叠则计数加一并将区间变成下一组，最后输出计数。

细节：保存区间时要用double类型储存。

#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>using namespace std;struct dao{    double left;    double right;};bool cmp(const dao &a,const dao &b){    if(a.left<=b.left) return true;    return false;}int main(){    dao ww[1000];    int n,d,i,l;    int x,y,t,k,e=1;    double L,R,a;    while(cin>>n>>d&&n!=0||d!=0)    {        t=1;l=1;        for(i=0;i<n;i++)        {            cin>>x>>y;            if(d>=y&&l)            {                ww[i].left=x-sqrt(d*d-y*y);                ww[i].right=x+sqrt(d*d-y*y);            }            else l=0;        }        if(!l)        {            cout<<"Case "<<e++<<": -1"<<endl;            continue;        }        sort(ww,ww+n,cmp);        R=ww[0].right;        for(k=1;k<n;k++)        {            if(ww[k].left>R)            {                t++;                R=ww[k].right;            }            else if(ww[k].right<=R)                {                    R=ww[k].right;                }        }        cout<<"Case "<<e<<": "<<t<<endl;        e++;    }    return 0;}