poj 1328 Radar Installation【贪心区间选点】

Radar Installation

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

Total Submission(s) : 22   Accepted Submission(s) : 9

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 2

1 2

-3 1

2 1

 

 

1 2

0 2

 

0 0

 

 

Sample Output

Case 1: 2

Case 2: 1

 

题解：先计算出可以覆盖小岛的区间，然后将这些区间按照区间右端从小到大排序，找出是否有重复区间

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define MAX 11000
struct node
{
	double beg;
	double end;
}s[MAX];
bool cmp(node a,node b)
{
	return a.beg<b.end;
}
int main()
{
	int i,j;
	int island,r;
	double a,b;
	int k=0;
    while(scanf("%d%d",&island,&r)&&island!=0&&r!=0)
    {
    	int ok=0;
    	for(i=0;i<island;i++)
    	{
    		scanf("%lf%lf",&a,&b);
    		if(fabs(b)>r)
    		{
    			ok=1;
    			break;
    		}
            s[i].beg=a-sqrt(r*r-b*b);//区间左端 
            s[i].end=a+sqrt(r*r-b*b);//区间右端 
    	}
    	if(ok)
    	{
    		printf("-1\n");
    		continue;
    	}
    	sort(s,s+island,cmp);
    	int sum=0;
    	double ans=-11000.0;
        for(i=0;i<island;i++)
        {
        	if(ans<s[i].beg)
        	{
        		ans=s[i].end;
        		sum++;
        	}
        }
        printf("Case %d: ",++k);
        printf("%d\n",sum);
    }
	return 0;
}