poj 1328 Radar Installation 贪心

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 123150		Accepted: 27237

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

题意：问最少需要多少个雷达才能覆盖所有小岛

题解：求出每个小岛对应一个雷达的安装区间[l,r],按r从小到大排序，若r相等，按l从大到小排序。

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
using namespace std;
struct node
{
    double l;
    double r;
}p[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,d,x,y,t=1,flag;;
    while(scanf("%d%d",&n,&d)&&n&&d)
    {
        flag=0;
        for(int i=0;i<n;i++)
        {
            cin>>x>>y;
            double k;
            k=sqrt((double)d*(double)d-(double)y*y);
            p[i].l=x-k;
            p[i].r=x+k;
            if(y>d||d<0)
                flag=1;
        }
        if(flag==1)
            printf("Case %d: -1\n",t++);
        else
        {
            sort(p,p+n,cmp);
            double m=p[0].r;
            int cnt=1;
            for(int i=1;i<n;i++)
            {
                if(p[i].l>m)
                {
                    cnt++;
                    m=p[i].r;
                }
            }
            printf("Case %d: %d\n",t++,cnt);
        }
        
    }
}