POJ-1328 Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 117779		Accepted: 25982

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

题目大意：以x轴为海岸线，上方为海，下方为陆地，海岸上的点表示雷达，海里的点表示岛屿，雷达的扫描区域为一个以该雷达为圆心半径为d的圆，问最少装多少个雷达可以使每个岛屿都被扫描到。

做题思路：可将问题转化为：求最少用多少个点可以使x轴上每个区间内都有一个点。

并且当y>d时，该岛屿没有满足的雷达。

AC代码：

#include 
#include 
#include 
#include 
using namespace std;
struct node
{
    double left;
    double right;
};
double compute(double p,double q)
{
    double t=p*p-q*q;
    return sqrt(t);
}
bool cmp(node aa,node bb)
{
    if(aa.rightd)
                flag=true;
            double x1=compute(d,y);
            a[i].left=x-x1;
            a[i].right=x+x1;
        }
        if(flag==true)
        {
            printf("Case %d: -1\n",++k);
            continue;
        }
        sort(a,a+n,cmp);//将结构体数组按照right从小到大排序
        int pos=0;//每次比较的位置
        summ=1;//初始为一个
        for(int i=1; i

如有不对，或者疏漏的点，还请各位大佬多多指教