POJ-1328Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 67259		Accepted: 15091

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

反向考虑，把雷达建立到岛上，则其圆形区域与x轴相交的弦上每个点都可以放雷达侦测到这个岛屿，同理，求其他所有岛屿，按弦的左边坐标排序，把雷达建立到右点上，如果下一段的左端点大于前一个的右端点，则需要一个新的雷达，且雷达建立在这一段弦的右端点，如果下一段的左端点小于前一个的右端点，则不需要新建雷达，但是雷达的位置要改变到下一段的右端点，如此重复！

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
struct node
{
double l,r;
}s[1100];
bool cmp(node x,node y)
{
return x.l<y.l;
}
int main()
{
int m,i,j,cot=0,flag;
double r,x,y;
while(scanf("%d%lf",&m,&r),m+r)
{
int sum=1;
flag=0;
for(i=0;i<m;i++)
{
scanf("%lf%lf",&x,&y);
s[i].l=x-sqrt(r*r-y*y);
s[i].r=x+sqrt(r*r-y*y);
if(y>r)
flag=1;
}
if(flag)
{
printf("Case %d: -1\n",++cot);
continue;
}
sort(s,s+m,cmp);
int t=0;
for(i=1;i<m;i++)
{
if(s[i].l>s[t].r)
{
sum++;
t=i;
}
else if(s[i].r<=s[t].r)
t=i;
}
printf("Case %d: %d\n",++cot,sum);
}
return 0;
}