NYOJ 287 Rader ( 区间选点问题）（有图解）

描述

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.

 

输入

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

输出

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.

样例输入

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

样例输出

Case 1: 2
Case 2: 1

来源

Beijing 2002

AC代码：

//这道题应该抽象为P1 P2 P3 。。。向X轴发射半径为r的波，例如P1为圆心画圆与X轴相交的两点之间便是雷达想要扫描到P1的活动范围，那么P2，P3。。。以此类推，它们与X轴分别相交的线段区间的公共区域，便是雷达想要同时扫描到它们时，雷达应该在的区域，就像线段竖向排，每次画一条竖线穿过能穿过的线段，画几条能将线段全部穿过，如下图示意：

//即区间选点问题： 数轴上有n个闭区间。取尽量少的点，使得每个区间内都至少有一个点。

//注意：变量要设成浮点型，并且雷达只会在X轴上移动

#include <stdio.h>

#include <math.h>
#include <algorithm>
using namespace std;
struct node 
{
double l,r; //要设成double型，否则会精度丢失，WA
}a[1005];
int cmp(node x,node y)
{
if(x.r<y.r) return 1;//让右端点小的排在前面
else if(x.r==y.r&&x.r<y.r) return 1;
return 0;
}
int main()
{
double x,y,r;
int n,count=1;
while(~scanf("%d %lf",&n,&r))
{
if(n==0&&r==0) return 0;
int flag=0;
for(int i=0;i<n;i++)
{
scanf("%lf %lf",&x,&y);
if(y>r) flag=1;//如果没有方案，即雷达能扫描的半径比它离X轴的距离小，雷达触碰不到
else
{
a[i].l=x-sqrt(r*r-y*y);
a[i].r=x+sqrt(r*r-y*y);
}
}
if(flag==1) printf("Case %d: -1\n",count);
else
{
sort(a,a+n,cmp);
   double max=a[0].r;
            int ans=1;
            for(int i=0;i<n;i++)
            {
           if(a[i].l>max) 
           {
           ans++;
           max=a[i].r;
   }
   }
   printf("Case %d: %d\n",count,ans);
}
count++;
} 
return 0;
}