poj1328radar installation 【贪心】

 

Radar Installation

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 93   Accepted Submission(s) : 46

Problem 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 21 2-3 12 11 20 20 0

 

Sample Output

Case 1: 2Case 2: 1
 
【解题思路】：以小岛为圆心，以d为半径作圆，圆与x轴的两个交点之间的部分即为雷达可以放置的位置，
左交点记为l，右交点记为r，比较区域重合的部分，与nyoj14【会场安排问题】、hdu2037【今年暑假不AC】
是同种类型。
 
 
 
【代码】
#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;#define maxn 1100struct radar{double l,r;}a[maxn];bool cmp(radar x,radar y){if(x.r==y.r)  return x.l<y.l;else   return x.r<y.r;}int main(){int n,d,i,x,y,flag,count;int k=1;double temp;while(scanf("%d%d",&n,&d)&&(n||d)){flag=0;for(i=0;i<n;i++){scanf("%d%d",&x,&y);if(y>d)    {    flag=1;    continue;    }    temp=sqrt(d*d*1.0-y*y);    a[i].l=x-temp;    a[i].r=x+temp;        }    printf("Case %d: ",k++);    if(flag)    {    printf("-1\n");    continue;}sort(a,a+n,cmp);count=1;temp=a[0].r;for(i=1;i<n;i++){if(a[i].l>temp){temp=a[i].r;count++;}  }printf("%d\n",count);}return 0;}