POJ1328

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

 

题意:每个雷达有个覆盖范围,求覆盖所有岛屿的最少雷达数

思路:贪心策略,为了使雷达数量最少,肯定是将雷达尽量设置在右边的位置。所以预处理所有的岛屿,求出覆盖每座岛屿的x坐标的取值范围,然后按最右边范围从小到大排序,依次选取和检查即可。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define INF 1<<28
static const int MAX = 1050;
int n, d;

struct island{
    int x, y;
    double l, r;    //每个点的左右边界

    //按范围的最大取值横坐标升序排序
    bool operator < (const struct island &p)
    {
        return r < p.r;
    }
}A[MAX];

//在岸上找一个点,该点越靠近右边越好
int main()
{
    int k = 1;
    while (scanf("%d %d", &n, &d) && n && d)
    {
        bool nosolution = false;
        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%d %d", &A[i].x, &A[i].y);
            if (A[i].y > d)
                nosolution = true;
        }
        if (nosolution)
            printf("Case %d: -1\n", k++);
        else
        {
            for (int i = 0; i < n; i++)
            {
                A[i].l = A[i].x - (double)sqrt((double)d * d - (double)A[i].y * A[i].y);
                A[i].r = A[i].x + (double)sqrt((double)d * d - (double)A[i].y * A[i].y);
            }
            sort(A, A + n);

            for (int i = 0; i < n; )
            {
                double p = A[i].r;
                while (p >= A[i].l && p <= A[i].r)
                    i++;
                cnt++;
            }
            printf("Case %d: %d\n", k++, cnt);
        }
    }
    return 0;
}







 

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