[POJ 1328] Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 59563		Accepted: 13430

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

 

区间选点问题

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
using namespace std;
#define N 1010

struct P
{
    double x1,x2;
    bool operator < (const P &t)const
    {
        if(x2!=t.x2) return x2<t.x2;
        return x1>t.x1;
    }
}p[N];

int main()
{
    int n,r;
    int iCase=1;
    int end_flag;
    while(scanf("%d%d",&n,&r),n||r)
    {
        end_flag=0;
        for(int i=1;i<=n;i++)
        {
            double x,y;
            scanf("%lf%lf",&x,&y);
            if(y>r) end_flag=1;
            p[i].x1=x-sqrt(r*r-y*y);
            p[i].x2=x+sqrt(r*r-y*y);
        }
        printf("Case %d: ",iCase++);
        if(end_flag)
        {
            printf("-1
");
            continue;
        }
        sort(p+1,p+n+1);
        int cnt=1;
        double tmp=p[1].x2;
        for(int i=2;i<=n;i++)
        {
            if(p[i].x1>tmp)
            {
                cnt++;
                tmp=p[i].x2;
            }
        }
        printf("%d
",cnt);
    }
    return 0;
}