（贪心，区间选点问题） Radar Installation POJ 1328

最新推荐文章于 2021-11-11 18:31:11 发布

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#贪心 #区间选点

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本文介绍了一种算法，用于确定最少数量的雷达安装位置以覆盖所有岛屿。通过将岛屿视为坐标系中的点，并考虑雷达覆盖范围，算法计算了每个岛屿的有效覆盖区间。在区间可能重叠的情况下，提出了解决方案来最小化所需雷达数量。

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

这道题思路就是把海岛当成圆心，然后算出每一个海岛的选点区间

算完每个海岛的选点区间后，如果有区间重叠，则说明可在重叠的部分放置一个鱼雷即可

下面是代码。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;
struct Node{
    double left;
    double right;
}node[1010];
bool cmp(const Node &p,const Node &q){
    return (p.right<q.right||(p.right==q.right&&p.left>q.left));
}
int main()
{
    int n,t=1;
    double d;
    while(cin>>n>>d&&(n+d))
    {
        int flag=0;
        int tx,ty;
        for(int i=1;i<=n;++i)
        {
            cin>>tx>>ty;
            if(ty>d)
                flag=1;
            node[i].left=tx-sqrt(d*d-ty*ty);
            node[i].right=tx+sqrt(d*d-ty*ty);
        }
        if(flag)
            cout<<"Case "<<t<<": "<<-1<<endl;
        else
        {
            sort(node+1,node+n+1,cmp);
            int num=0;
            double tright=-INF;
            for(int i=1;i<=n;++i)
            {
                if(node[i].left>tright)
                {
                    num++;
                    tright=node[i].right;
                }
            }
            cout<<"Case "<<t<<": ";
            cout<<num<<endl;
        }
        ++t;
    }
    return 0;
}