E - Radar Installation

标签：efi   ons   def   using   air   inpu   over   ide   安装   

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

想要按照最少的雷达站，先把x排序，如何判断下一个的最左边是不是在之前的雷达的最右边的右边，是就要重新按一个雷达站；

要注意，如果下一个的最右边在之前雷达的左边，说明安装的位置要在这里的最右边；

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define cl clear()
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
const double pi=acos(-1.0);
typedef __int64 ll;
typedef long double ld;
const ll mod=1e9+7;
using namespace std;
int n,d,bits=1;
double wzl(int y)
{
    double x=d*d-y*y;
    return sqrt(x);
}
int a[1005],b[1005];
int main()
{
    
    while(1)
    {
        mm(a,0);
        mm(b,0);
        sf("%d%d",&n,&d);
        if(n==0)
        return 0;
        int temp=0;
        for(int i=0;i<n;i++)
        {
            sf("%d%d",&a[i] ,&b[i] );   
            if(b[i]>d) 
            temp=1;
        }
        if(temp) 
        {
            pf("Case %d: -1\n",bits++);
            continue;
        }
        for(int i=0;i<n-1;i++)
        for(int j=0;j<n-i-1;j++)
        {
            if(a[j]>a[j+1])
            {
                int w=a[j];
                a[j]=a[j+1];
                a[j+1]=w;
                w=b[j];
                b[j]=b[j+1];
                b[j+1]=w;
            }
        }
        int sum=1;
        double last=a[0]+wzl(b[0]),left,right;
        for(int i=1;i<n;i++)
        {
            left=a[i]-wzl(b[i]);
            right=a[i]+wzl(b[i]);
            if(left>last)
            {
                sum++;
                last=right;
            }else if(right<last)
            {
                last=right;
            }
        }
        pf("Case %d: %d\n",bits++,sum);
    }
}

E - Radar Installation

标签：efi   ons   def   using   air   inpu   over   ide   安装   

原文地址：https://www.cnblogs.com/wzl19981116/p/9350032.html