由於poj不能使用c11，改在牛客網上OJ

題目地址

題解

以每個島為中心，d為半徑畫個圓，交x軸為兩點，轉化為活動安排問題，即求最多有多少不覆蓋的線段。

程式碼

#include <algorithm>
#include <cmath>
#include <iostream>
#include <vector>
using namespace std;
const double INF = 1e9;
struct node {
    double s, e;
};
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, d;
    while (cin >> n >> d, n != 0 && d != 0) {
        vector<node> record(n);
        bool noSolution = false;
        for (auto &i : record) {
            double t0, t1;
            cin >> t0 >> t1;
            double tmp = d * d - t1 * t1;
            if (tmp < 0) {
                noSolution = true;
                continue;
            }
            double t2 = sqrt(d * d - t1 * t1);
            i.s = t0 - t2, i.e = t0 + t2;
        }
        if (noSolution) {
            cout << -1 << '\n';
            continue;
        }
        sort(record.begin(), record.end(),
             [](node &a, node &b) { return a.e < b.e; });
        int cnt = 0;
        double lastEnd = -INF;
        for (auto &i : record) {
            if (i.s > lastEnd) {
                ++cnt;
                lastEnd = i.e;
            }
        }
        cout << cnt << '\n';
    }
    return 0;
}