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POJ1328 Radar Installation

  • 题目链接 POJ1328

  • 题目描述

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.


poj1328
poj1328
  • 输入

    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

  • 输出

    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.

  • 样例输入

3 2
1 2
-3 1

2 1
1 2
0 2

0 0
  • 样例输出

    Case 1: 2
    Case 2: 1

  • 思路

    由于所有雷达全在x轴上,可把每个岛屿以d为圆心画圆,记录圆与x轴的两个交点存入pair中,把所有点对按右交点升序排列,判断后一个左交点是否小于前一个右交点即可。

  • 代码

#include<iostream>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;

vector<pair<double,double> > v;

int main()
{
    int n,d,cnt=0;
    while(cin>>n>>d)
    {
        cnt++;
        if(n==0&&d==0)
            return 0;
        int maxx=-1;
        for(int i=0;i<n;i++)
        {
            int x,y;
            cin>>x>>y;
            v.push_back(make_pair(x+sqrt(abs(1.0*d*d-y*y)),x-sqrt(abs(1.0*d*d-y*y))));
            maxx=max(maxx,y);
        }
        if(maxx>d)
        {
            cout<<"Case "<<cnt<<": -1"<<endl;
            continue;
        }
        sort(v.begin(),v.end());
        int ans=1;
        double tmp=v[0].first;
        for(int i=1;i<n;i++)
        {
            if(v[i].second>tmp)
            {
                ans++;
                tmp=v[i].first;
            }
            else if(v[i].first<tmp)
                tmp=v[i].first;
        }
        cout<<"Case "<<cnt<<": "<<ans<<endl;
        v.clear();
    }
}

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