Radar Installation

Radar Installation
Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu
Submit Status

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

雷达可以位于岛屿的左侧也可以位于雷达的右侧。而这就可以分别确定雷达相对与岛屿x的最左坐标和x最右坐标。
    最左为：x - sqrt(d*d-y*y); 最右为：x + sqrt(d*d-y*y); 
    每个岛屿都有这样的最左和最右可被侦测坐标。
    根据贪婪的思想，每次都应该将最右可被侦测坐标作为衡量标准。
假定当前的岛屿为cur，当前的下一个为next。
    如果next的最左可被侦测坐标比cur的最右都大的话，只能再设一个雷达来侦测next了。（转）

解题思路：将二维坐标转换为一维坐标上的区间。求最小点覆盖所有区间。如果小岛能被雷达覆盖，则转换在一维坐标上的最大区间为，[x-sqrt(d*d*1.0-y*y),x+sqrt(d*d*1.0-y*y)]
如果其中某个小岛不能被覆盖，则用标记变量标记。直接输出-1。如果所有小岛都能被雷达覆盖，则进行贪心策略。
对所有区间右端进行从小大到排序（右端相同时，左端从大到小排序），则如果出现区间包含的情况，小区间一定排在前面。
然后开始遍历区间。如果当前区间超过上一个区间的覆盖范围，则雷达数+1。第一个区间去最右值。
证明：如果第一个区间不取最右值，而取中间的点，那么把点移到最右端，被满足的区间增加了。而原先被满足的区间现在一定被满足。（转） 代码：#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
double s,e;
}a[1010];
int cmp(node x,node y)
{
if(x.e!=y.e)
return x.e<y.e;
else
return x.s>y.s;
}
int main()
{
int n,d,i,j,k=1;
double temp;
while(scanf("%d%d",&n,&d)&&(n||d))
{
int x,y;
int 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].s=x-temp;
a[i].e=x+temp;
}
printf("Case %d: ",k++);
if(flag)
{
printf("-1\n");
continue;
}
sort(a,a+n,cmp);
int ans=1;
temp=a[0].e;
for(int i=1;i<n;i++)
{
if(a[i].s>temp)//如果next的最左可被侦测坐标比cur的最右都大的话，
//只能再设一个雷达来侦测next了。
{
ans++;
temp=a[i].e;
}
}
printf("%d\n",ans);
}
return 0;
}