POJ 1328 Radar Installation(区间选点)

　　　　　　　　　　　　　　　　　　Radar Installation

 

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

根据半径和点的纵坐标求出圆心的坐标范围，形成一系列区间，然后就是区间选点的问题了。

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

struct len
{
    double start;
    double end;
};

bool cmp(len a,len b)
{
    if(a.end==b.end)
    return a.start<b.start;
    return a.end<b.end;
}

double cal(int y,int d)
{
    return sqrt((double)(d*d-y*y));
}

int main()
{
    //freopen("in.txt","r",stdin);
    bool flag;
    len a[1005];
    int n,d,i;
    int kase=0;
    while(scanf("%d%d",&n,&d),n||d)
    {
        flag=false;
        for(i=0;i<n;i++)
        {
            int  x,y;
            scanf("%d%d",&x,&y);
            if(flag)
            continue;
            if(y>d)
            flag=true;
            double temp=cal(y,d);
            a[i].start=x-temp;
            a[i].end=x+temp;
        }
        if(flag)
        printf("Case %d: -1\n",++kase);
        if(!flag)
        {
            sort(a,a+n,cmp);
            int cnt=1;
            double v=a[0].end;
            for(i=1;i<n;i++)
            {
                if(v<a[i].start)
                {
                    cnt++;
                    v=a[i].end;
                }
            }
            printf("Case %d: %d\n",++kase,cnt);
        }
    }
    return 0;
}