CSU-ACM2017暑期训练7-模拟&&贪心 A - Radar Installation POJ - 1328

题目：

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

题意：x轴以上为海洋，x轴及以下为陆地，海洋中有若干个小岛，现想要侦测到他们，给你一种雷达的检测半径，雷达只能放在陆地上，问最少需要多少个雷达能覆盖全所有的小岛。

思路：显然雷达全放在x轴上是离海洋最近的，对于每一个小岛，都有一个在x轴上放置雷达能覆盖到他的区间，求出这个区间的两端，然后贪心求解即可，具体看代码。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn = 100;
int n,d;
int ok,no=0;

struct Island
{
double l,r;
}island[1111];

bool cmp(Island a,Island b)
{
return a.l<b.l;
}

int main()
{
while(scanf("%d%d",&n,&d)!=EOF)
{
if(n==0&&d==0)
break;
ok=1;
for(int i=0;i<n;i++)
{
int a,b;
scanf("%d%d",&a,&b);
island[i].l=(double)a-sqrt((double)d*d-b*b);
island[i].r=(double)a+sqrt((double)d*d-b*b);
if(b>d)
ok=0;
}
if(!ok)
{
printf("Case %d: -1\n",++no);
continue;
}
sort(island,island+n,cmp);
int ans=1;
double radar_x=island[0].r;
for(int i=1;i<n;i++)
{
if(radar_x>island[i].r)
{
radar_x=island[i].r;
}
else if(radar_x<island[i].l)
{
radar_x=island[i].r;
ans++;
}

}
printf("Case %d: %d\n",++no,ans);
}
return 0;
}