POJ 1328-Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 56977		Accepted: 12847

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

小结：

   新的弱项，gets...贪心也不好...

以下是AC代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
    double l,r;
}point[1500];
int compare(struct node a,struct node b)
{
    return a.l<b.l;
}
int main()
{
    int n,m,t=0;
    double x,y;
    while(scanf("%d%d",&n,&m)!=EOF&&(m+n))
    {
        int ans = 1;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf %lf", &x, &y);
            if(y > m)
                ans = -1;
            point[i].l = x - sqrt(m*m - y*y);
            point[i].r = x + sqrt(m*m - y*y);
        }
        if(ans == -1)
        {
            printf("Case %d: -1\n", ++t);
            continue;
        }
        sort(point,point+n,compare);
        double temp=point[0].r;


        for(int i=1;i<n;i++)
        {
            if(point[i].l>temp)
            {
                ans++;
                temp=point[i].r;
            }
            else if(point[i].r<temp)
            {
                temp=point[i].r;
            }
        }
        printf("Case %d: %d\n",++t,ans);
    }
    return 0;
}