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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.
Input The input consists of several test cases. The first line of each case contains two integers n (<=n<=) 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. 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 - Sample Output Case : Source |
Figure A Sample Input of Radar Installations贪心。
一开始的思路：
图中ABCD为海岛的位置。假设本题半径为2（符合坐标系），那么A点坐标为（1，1）以此类推。
在题中，记录每个点的坐标，并把一个点新加一个标记变量以标记是否被访问过。
首先以A为圆心，r为半径做圆，交x轴与E1（右）点，做出如图中绿色虚线圆。然后以E1为圆心，半径为r做圆。看此时下一点（B）是否在该圆中。如果在，那么将该点标记变量变为true，再判下一点（C），如果不在那么就新增一个雷达。
后来，换了思路，存储图中红色圆与X轴的交点。
还是原来的贪心思路，仍然先排序，但排序的准则是按红色圆与x轴左交点的先后顺序。
如图，如果B点圆（且如此称呼）的左交点（J1点）在A点圆右交点（E1）左侧，那么B点一定涵盖在A点圆内部。再如图，如果D点圆的左交点（图中未标示）在A点圆的右交点（未标示）右，则D点不在A点圆中。
故，有如下代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<cmath>
using namespace std;
struct Node
{
double left,right;
}p[];
bool cmp(Node a,Node b)
{
return a.left<b.left;
}
int main()
{
int n;
double r;
int ac=;
while(scanf("%d%lf",&n,&r)== && (n || r))
{
int f=;
for(int i=;i<n;i++)
{
double a,b;
scanf("%lf%lf",&a,&b);
if(b>r)
{
f=;
}
else
{
p[i].left=a-sqrt(r*r-b*b);
p[i].right=a+sqrt(r*r-b*b);
}
}
printf("Case %d: ",++ac);
if(f==)
{
printf("-1\n");
continue;
}
sort(p,p+n,cmp);
int ans=;
Node tmp=p[];
for(int i=;i<n;i++)
{
if(p[i].left>tmp.right)
{
ans++;
tmp=p[i];
}
else if(p[i].right<tmp.right)
{
tmp=p[i];
}
}
printf("%d\n",ans); }
return ;
}