Radar Installation

1328 poj

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 58386		Accepted: 13155

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

 

先将之转化为各区间 接下来就是区间选点问题了 排序后选最后一个点就OK；

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
struct node{
    double x;
    double y;
    double l;
    double r;
}island[1005];

bool cmp(node a,node b)
{
    if(a.r!=b.r)
        return a.r<b.r;
    else
        return a.l>b.l;
}

int main()
{
    int n,i,j,k,t=0,flg,num;
    double d,local;
    while(scanf("%d %lf",&n,&d)!=EOF)
    {
        flg=1;t++;num=1;
        if(n==0 && d==0)
        {
            break;
        }
        for(i=1;i<=n;i++)
        {
            scanf("%lf %lf",&island[i].x,&island[i].y);
            if(island[i].y>d)
                flg=0;
        }
        if(flg==0)
        {
            printf("Case %d: -1
",t);
            continue;
        }
        for(i=1;i<=n;i++)
        {
            island[i].l=island[i].x-sqrt(d*d-island[i].y*island[i].y);
            island[i].r=island[i].x+sqrt(d*d-island[i].y*island[i].y);
        }
        sort(island+1,island+n+1,cmp);
        local=island[1].r;
        for(i=2;i<=n;i++)
        {
            if(island[i].l<=local)
                continue;
            else
            {
                num++;
                local=island[i].r;
            }
        }
        printf("Case %d: %d
",t,num);
    }
    return 0;
}