△POJ1328--Radar Installation（贪心）

Radar Installation

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（转）：是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达（在x轴上），则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)]。

这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内斗至少存在一个点。每次迭代对于第一个区间， 选择最右边一个点， 因为它可以让较多区间得到满足，。如果不选择第一个区间最右一个点（而选择前面的点）， 那么把它换成（前面区间）最右的点之后， 以前得到满足的区间， 现在仍然得到满足。

 

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

#define maxn 1001

struct node
{
    double left;
    double right;
}island[maxn];

bool cmp(node a,node b)
{
    return a.left<b.left;
}

int main()
{
    int i,n,k=1,count,flag;
    double d,x,y,temp;
    while(scanf("%d%lf",&n,&d) != EOF)
    {
        flag=0;
        if(n==0&&d==0)
            break;
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf",&x,&y);
            if(y>d||d<0)
            {
                flag=1;
            }
            island[i].left=x-sqrt(d*d-y*y);//先算出每个岛的雷达覆盖区间
            island[i].right=x+sqrt(d*d-y*y);//先算出每个岛的雷达覆盖区间
        }
        if(flag)
        {
            printf("Case %d: -1
",k++);
            continue;
        }
        sort(island,island+n,cmp);//按左端点排序
        temp=island[0].right;//排序完成后，第一个雷达建立在第一个区间的右端点
        count=1;
        for(i=1;i<n;i++)
        {//如果左端点在雷达左边，这个时候要考虑区间的右端在雷达的左边还是右边，如果是右边，那雷达位置就不变
            if(island[i].right<temp)//如果在左边，那现在的雷达是覆盖不了的
            {
                temp=island[i].right;//所以要把雷达放在该区间的右端点，这样能同时覆盖原来的岛和现在的岛
            }
            else if(island[i].left>temp)//判断每个区间的左端点，如果在最新建立的雷达右边
            {
                count++;//那么肯定需要一个新雷达
                temp=island[i].right;//而且也建在该区间右端
            }
        }
        printf("Case %d: %d
",k++,count);
    }
    return 0;
}