poj 1328 Radar Installation(贪心)

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\n",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\n",coun++,ans);
    }
    return 0;
}