Radar Installation -poj 1328

 

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 60277		Accepted: 13583

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
使用贪心算法解决。

#include <iostream>
#include<math.h>
#include <algorithm>
#include <stdlib.h>
using namespace std;
struct point
{
    double left, right;
}points[1000], temp;

bool operator < (point a, point b)
{
    return a.left < b.left;
}
int main() {
    int num = 0;
    double rad = 0;
    cin >> num >> rad;
    int casenum=1;
    while (num || rad) {


        int rnum=1;
        int wrong = 0;

        for (int i = 0; i < num; i++) {
            double x=0;
            double y=0;
            cin >> x >> y;

            if (y > rad) {
                //岛屿纵坐标大于雷达半径
                wrong = 1;
            }else{
                //以岛屿为圆心，做圆，与x轴的交点
                points[i].left=x-sqrt((rad*rad-y*y)*1.0);
                points[i].right=x+sqrt((rad*rad-y*y)*1.0);

            }
        }
        if (wrong) {
            //雷达不能覆盖
            cout<<"Case "<<casenum++<<": "<<-1<<endl;
        } else {
            //按照岛屿画圆与x轴的左交点排序
            sort(points, points + num);
             temp=points[0];
            for(int i=1;i<num;i++){
                //
                if(points[i].left>temp.right)//temp.right为公共区间的右端点，若下一个区间的左端点大于temp.right则该区间与以上的公共区间没有公共部分
                {
                    rnum++;//雷达个数加1
                    temp=points[i];//更新公共区间右端点
                }else if(points[i].right<temp.right){
                     temp=points[i];//更新公共区间右端点
                }
                }

            cout<<"Case "<<casenum++<<": "<<rnum<<endl;
        }

        cin >> num >> rad;
    }
    return 0;
}