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
首先计算出每个海岛对应的能够在海岸线上修建雷达站的区间
再按区间的左值排序 题目就转化为了区间选点的贪心问题
先设定一个能够架设当前雷达的区间
当岛屿的区间左值大于雷达区间右值时 雷达数加一 并设定下一个雷达区间
当岛屿的区间右值小于雷达区间右值时 雷达区间右值向左缩至当前岛屿区间
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll __int64
#define MAXN 1000
#define INF 0x7ffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
struct Island{
double l,r;
};
Island s[MAXN+10];
int cmp(Island a,Island b)
{
return a.l<b.l;
}
int main()
{
int n,coun=1,ans;
double d,x,y;
int i,j,ok;
while(cin>>n>>d)
{
if(n==0&&d==0) break;
ok=1;ans=0;
for(i=1;i<=n;i++)
{
scanf("%lf%lf",&x,&y);
if(y>d) ok=0;
if(ok)
{
double l=sqrt(d*d-y*y);
s[i].l=x-l;
s[i].r=x+l;
}
}
if(!ok) {printf("Case %d: -1
",coun++);continue;}
sort(s+1,s+n+1,cmp);
double left=-INF*1.0,right=-INF*1.0;
for(i=1;i<=n;i++)
{
if(s[i].l>right)
{
ans++;
left=s[i].l;
right=s[i].r;
}
else if(s[i].r<right) right=s[i].r;
}
printf("Case %d: %d
",coun++,ans);
}
return 0;
}