POJ 1328 区间选点问题

http://poj.org/problem?id=1328

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

网上的详解

我只说一句话：注意文章中那些变量一定要定义成double

我的代码：

#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <iostream>
using namespace std;
struct A
{
    double l,r;
};
int cmp(const A &a,const A &b)
{
    if(a.r==b.r)
        return a.l>b.l;
    return a.r < b.r;
}
int main()
{
    int n,d,t=1;
    while(~scanf("%d%d",&n,&d))
    {
        if(n==0&&d==0)
          break;
        A p[1005];
        int flag=1;
        int x,y;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&x,&y);
            if(y<=d)
            {
                p[i].l=x-sqrt((double)d*d-(double)y*y);
                p[i].r=x+sqrt((double)d*d-(double)y*y);
            }
            else
                flag=0;
        }
        if(!flag)
        {
            cout<<"Case "<<t++<<": "<<-1<<endl;
            continue;
        }
        sort(p,p+n,cmp);
        int sum=1;
        double end=p[0].r;
        for(int i=1; i<n; i++)
        {
            if(end<p[i].l)
            {
                end=p[i].r;
                sum++;
            }
        }
        printf("Case %d: %d\n",t++,sum);
    }
    return 0;
}