POJ1328 -- Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 96747		Accepted: 21508

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

Source

Beijing 2002

 

题意分析
给定一个直角坐标系，定义x轴为海岸线，海岸线上方是海，下方是陆地.
在海域零星分布一些海岛, 需要要在海岸线上安装若干个雷达覆盖这些海岛,
每个雷达的覆盖半径都是相同且固定的.

现在给定所有海岛的坐标(x,y), 以及雷达的覆盖半径d，
问可以覆盖所有海岛的最小雷达数.

解题思路：

无解的情况：

　　y>d或者是y<0或是d<0

有解的情况：

以海岛坐标(x,y)为圆心，用雷达覆盖半径d画圆，根据前提条件可知，
这个圆与x轴必定存在最少1个(y=d)、最多2个交点(y<d).

可以认为1个交点是2个交点重合的特殊情况，那么这2个交点在x轴上构成的线性区间，
可以看作海岛的被覆盖范围在x轴上的投影 (由此就可以把二维的几何问题转化成一维几何问题)

按照所有海岛的x轴坐标，从小到大依次计算每个海岛在x轴上的投影区间（投影可能存在重叠），
同时把每个雷达抽象成1个点，那么此问题就转化成：

【已知x轴上若干个区间（可能存在交集），现在要往这些区间内放若干个点，
问怎样放置这些点，使得每个区间内至少存在一个点，且所放置的点的总量尽可能最少】


可使用贪心算法求解：
根据每个区间的左端点坐标对所有区间从左到右排序:
① 在左起第一个区间A 的右端点 A.R 放置一个点，总放置点数 P+1
② 若下一个区间B 的左端点 B.L > A.R,
说明 区间A 与 区间B 无交集，此时在区间B 的右端点 B.R 放置一个点，总放置点数 P+1

否则说明 区间A 与 区间B 相交, 此时进一步判断，若 B.R < A.R，
说明 区间B 在 区间A 的内部，此时把原本放置在 A.R 的点移动到 B.R（确保区间B有一个点），总放置点数不变

③ 重复这个过程直到所有区间被遍历完成

测试样例：
3 2
1 0
-3 -1
2 1

2 2
1 1
2 2

5 4
-2 2
1 2
0 3
3 3
4 3

9 3
0 2
-3 2
-4 2
-5 2
3 2
6 2
9 2
12 2
15 2

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
struct seaside{
    double left;
    double right;
};
bool compare(seaside a,seaside b)
{
    return a.left<=b.left;
}
int solve(seaside *a,int n)
{
    ///进行排序
    sort(a,a+n,compare);
    double fright;
    fright = a[0].right;
    int sum = 1;
    for(int i=1;i<n;i++)
    {
        if(a[i].left > fright)
        {
            sum++;
            fright = a[i].right;
        }else{
            if(a[i].right < fright)
            {
                fright = a[i].right;
            }
        }
    }
    return sum;
}
int main()
{
    int n;
    double d;
    int Count = 1;
    bool isSolved = 1;
    cin>>n>>d;
    while(n!=0)
    {
        isSolved = 1;
        seaside *island = new seaside[n];
        double x,y;
        for(int i=0;i<n;i++)
        {
            cin>>x>>y;
            if(y>d || y<0 || d<0)//无解
                isSolved = 0;
            else{
                double offset = sqrt(d*d - y*y);   // 勾股定理
                island[i].left = x-offset;
                island[i].right = x+offset;
            }

        }

        ///解决问题
        cout<<"Case "<<Count<<": ";
        if(isSolved)
            cout<<solve(island,n)<<endl;
        else
            cout<<-1<<endl;

        delete[] island;
        Count++;
        cin>>n>>d;
    }
    return 0;
}

结果：
3 2
1 0
-3 -1
2 1
Case 1: -1

2 2
1 1
2 2
Case 2: 1

5 4
-2 2
1 2
0 3
3 3
4 3
Case 3: 1

9 3
0 2
-3 2
-4 2
-5 2
3 2
6 2
9 2
12 2
15 2
Case 4: 4

3 2
1 2
-3 1
2 1
Case 5: 2

1 2
0 2
Case 6: 1

0 0

Process returned 0 (0x0)   execution time : 1.058 s
Press any key to continue.