Radar Installation 区间问题

最新推荐文章于 2021-11-11 18:31:11 发布

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本文介绍了一种算法，用于解决沿海岸线安装雷达以覆盖海上岛屿的问题。通过计算每个岛屿所需的雷达覆盖范围，并采用排序和贪婪策略，确定最少数量的雷达安装位置。

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A - Radar Installation

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

Submit Status

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

Sample Output

Case 1: 2
Case 2: 1

以岛屿为圆心，以雷达所能那个扫射到的距离为半径画圆交x轴，交点为node[i].left,node[i].right，对node进行排序，left小的排在前面

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define  MAX 1004


using namespace std;
int x[MAX],y[MAX];
typedef struct{
    double left,right;
}radar;


bool cmp(radar a, radar b)
{
    return a.left<b.left;
}
int main()
{
    radar node[MAX];
    int n,d,i,temp,t=1,count;
    double a,high;
    while(cin>>n>>d && (n || d))
    {
        temp=0,count=1;
        if(d<=0)
            temp=1;
        for(int i=0;i<n;i++)
        {
            cin>>x[i]>>y[i];
            if(y[i]>d) temp=1;
        }
        if(temp || n==0)
        {
            printf("Case %d: -1\n",t++);
            continue;
        }
        for(i=0;i<n;i++)
        {
            a=sqrt(d*d-y[i]*y[i]);
            node[i].left=x[i]-a;
            node[i].right=x[i]+a;
        }
        sort(node,node+n,cmp);
        high=node[0].right;
        for(i=1;i<n;i++)
        {
            if(node[i].left<=high)
            {
                high=node[i].right>high?high:node[i].right;
            }
            else
            {
                count++;
                high=node[i].right;
            }
        }
        printf("Case %d: %d\n",t++,count);
    
    }
    return 0;
    
}