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Radar Installation

时间:2023-01-07 01:00:19浏览次数:56  
标签:std 选点 Installation int Radar 区间 include

Radar Installation

题解:区间选点问题

为什么是区间选点呢?关键是我们反着来看,如果将岛屿看成圆心,我们以雷达照射范围d为半径,我们可以X轴上找到一段区间,这段区间都可以放雷达使得该岛屿能被照射到,所以一个个雷达就产生了一个个区间,让雷达数最少覆盖所有区间,不就是区间选点嘛

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
#define Zeoy std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0)
#define all(x) (x).begin(), (x).end()
#define endl '\n'
using namespace std;
const int mod = 1e9 + 7;
const double eps = 1e-9;
struct node
{
    double l, r;
    bool operator<(const node &t) const
    {
        return r < t.r;
    }
} a[1010];
int main(void)
{
    // Zeoy;
    int t = 1;
    // cin >> t;
    while (t--)
    {
        int cnt = 1;
        int n, d;
        while (cin >> n >> d)
        {
            if (!n && !d)
                break;
            printf("Case %d: ", cnt++);
            int flag = 1;
            for (int i = 1; i <= n; ++i)
            {
                double x, y;
                cin >> x >> y;
                a[i].l = x - sqrt(d * d - y * y);
                a[i].r = x + sqrt(d * d - y * y);
                if (y > d)
                {
                    flag = 0;
                }
            }
            if (!flag)
            {
                cout << "-1\n";
                continue;
            }
            sort(a + 1, a + n + 1);
            double now = a[1].r;
            int ans = 1;
            for (int i = 2; i <= n; ++i)
            {
                if (now < a[i].l)
                {
                    now = a[i].r;
                    ans++;
                }
            }
            cout << ans << endl;
        }
    }
    return 0;
}

标签:std,选点,Installation,int,Radar,区间,include
From: https://www.cnblogs.com/Zeoy-kkk/p/17032025.html

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