POJ 1328

逆向思维，假设岛屿是雷达，映射到\(x\) 轴的区间取交集，放置一个雷达即可

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
using namespace std;
const int N = 1e3 + 10;
struct node {
    double l,r;
}c[N];
bool cmp(node a,node b) {
    if(a.l != b.l) return a.l < b.l;
    return b.r < a.r;
}
int main() {
    int cnt = 0,n,d;
    while(cin >> n >> d && n && d) {
        bool ok = 1;
        for(int i = 0;i < n; ++i) {
            int x,y;
            cin >> x >> y;
            if(y > d) ok = 0;
            double dx = sqrt(d * d * 1.0 - y * y);
            c[i].l = x - dx;
            c[i].r = x + dx;
        }
        if(ok == 0) {
            cout << "Case " << ++cnt << ": -1" << endl;
            continue;
        }
        sort(c,c + n,cmp);
        double r = c[0].r;
        int ans = 1;
        for(int i = 0;i < n; ++i) {
            if(c[i].l <= r) r = min(r,c[i].r);
            else r = c[i].r,ans ++;
        }
        cout << "Case " << ++cnt << ": " << ans << endl;
    }
    return 0;
}

参考

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