Radar Installation（POJ 1328 区间贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 68578		Accepted: 15368

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

需要判断d<0,a[i].y>d情况。
首先，按照x坐标排序，对于每个岛屿求出雷达所能放置的区间，然后对这些进行处理，x1,x2;
设当前雷达放置位置为nowx,对于下一个区间，如果写x1>nowx,显然多需要一个雷达，反之如果nowx>x1,nowx=min(nowx,x2);

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;
 struct node
 {
     int x,y;
 }a[+];
 bool cmp(node q,node p)
 {
     if(q.x==p.x)
         return q.y>=p.y;
     return q.x<p.x;
 }
 int main()
 {
     int n,d;
     int i,j;
     int k=;
     freopen("in.txt","r",stdin);
     while(scanf("%d%d",&n,&d))
     {
         int coun=;
         if(n==&&d==)
             break;
         bool flag=;
         for(i=;i<n;i++)
         {
             scanf("%d%d",&a[i].x,&a[i].y);
             if(a[i].y>d)
                 flag=;
         }
         if(flag||d<=)
         {
             printf("Case %d: -1\n",k++);
             continue;
         }
         sort(a,a+n,cmp);
         double nowx=sqrt(double(d*d-a[].y*a[].y))+a[].x;
         double x1,x2,temp;
         for(i=;i<n;i++)
         {
             temp=sqrt(double(d*d-a[i].y*a[i].y));
             x1=a[i].x-temp;
             x2=a[i].x+temp;
             if(x1>nowx)
             {
                 nowx=x2;
                 coun++;
             }
             else if(nowx>x2)
                 nowx=x2;
         }
         printf("Case %d: %d\n",k++,coun);
     }
 }