Poj1328--Radar Installation(区间选点)

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 64121		Accepted: 14418

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

区间选点(转化思想)：
   <1>: 对左端点排序（小 → 大）， 左端点相同，（小→大）排右端点;
   <2>:对于num[0],在右端点放雷达，如果下一个区间左端点>现在右端点， 雷达shu+1, 如果左端点<右端点： 如果下一区间右端点<现在右端点， 更新雷达到下一区间右端点;

　

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

struct island
{
    double l, r; 
} num[1010];

bool cmp(island l, island r)
{
    if(l.l == r.l)
        return l.r < r.r;
    return l.l < r.l; 
}

int main()
{
    double r, x, y;    int i, m, t=1;
    while(~scanf("%d %lf", &m, &r))
    {
        int flag = 0;
        if(m == 0 && r == 0)
            break;
        for(i=0; i<m; i++)
        {
            scanf("%lf %lf", &x, &y);
            if(y > r)
            {flag = 1; continue; }
            num[i].l = x - sqrt(r*r - y*y);   //转化为区间问题； 
            num[i].r = x + sqrt(r*r - y*y);
        }
        if(flag)
        {
            printf("Case %d: -1
",t++); 
            continue;
        }
        sort(num, num+m, cmp);
        double temp = num[0].r; int total = 1;
        for(i=1; i<m; i++)        
        {
            if(num[i].l > temp){
                total++;
                temp = num[i].r;
            }
                
            if(num[i].l <= temp)
            {
                if(num[i].r < temp)
                    temp = num[i].r;
            }
        }
        printf("Case %d: %d
", t++, total); 
    }
    return 0;
}