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.
Input
The input consists of several test cases. The first line of each case contains two integers
n(1
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
題意：
題意：假設海岸線是一條無限延伸的直線，陸地在海岸線的一側，而海洋在另一側。每一個小的島嶼是海洋上的一個點，雷達坐落於海岸線上，只能覆蓋d距離，所以如果小島能夠被覆蓋到的話，它們之間的距離最多爲d，題目要求計算出能夠覆蓋給出的所有島嶼的最少雷達數目。果不能覆蓋，輸出-1。
思路：我們可以以點來做半徑爲d的圓，與x軸的相交，如果不相交那麼肯定完不成任務，反之就轉化成了區間選點問題。
代碼：
<span style="font-size:14px;">#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>//sort所在的庫文件，排序用
using namespace std;
const int MAXN=1005;
struct Line
{
double l,r;
}line[MAXN];//每個島作半徑爲d的圓，與x軸所截的線段
bool cmp(Line a,Line b)
{
return a.l<b.l;
} //按照線段的左端點從小到大排序
int main()
{
//freopen("test.in","r",stdin);
//freopen("test.out","w",stdout);
int n,d;
int i;
int x,y;
bool yes;//確定是不是有解
int icase=1;
while(cin>>n>>d)
{
yes=true;
int cnt=0;
if(n==0&&d==0)break;
for(i=0;i<n;i++)
{
cin>>x>>y;
if(yes==false)continue;
if(y>d)yes=false;
else
{
line[i].l=(double)x-sqrt((double)d*d-y*y);
line[i].r=(double)x+sqrt((double)d*d-y*y);
}
}
if(yes==false)
{
cout<<"Case "<<icase++<<": -1"<<endl;
continue;
}
sort(line,line+n,cmp);
cnt++;
double now=line[0].r;
for(i=1;i<n;i++)
{
if(line[i].r<now)
now=line[i].r;
else if(now<line[i].l)
{
now=line[i].r;
cnt++;
}
}
cout<<"Case "<<icase++<<": "<<cnt<<endl;
}
return 0;
}</span>
