题解三四条，细节第一条。 若是没有赞，作者两行泪！！！

#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;

const int MAXN = 15005;
int T;
int N, L, W;
struct node {
	double l, r;
}P[MAXN];
int cnt;

int main()
{
	while (scanf("%d", &T) != EOF)
	{
		for (int i = 1; i <= T; i++)
		{
			scanf("%d %d %d", &N, &L, &W);
			cnt = 0;
			for (int j = 1; j <= N; j++)
			{
				double C, R;
				scanf("%lf %lf", &C, &R);
				double x = 1.0 * sqrt((1.0 * R * R) - ((W / 2.0) * (W / 2.0))); 
				if (R * 2 <= W) continue;
				++cnt;
				P[cnt].l = max((double)0, C - x); 
				P[cnt].r = min((double)L, C + x); 
			}
			int ans = 0;
			double maxr = 0.0;
			bool check = 0;
			while (maxr < L)
			{
				double nowl = maxr;
				for (int j = 1; j <= cnt; j++)
					if (P[j].l <= nowl && P[j].r > maxr) maxr = P[j].r;
				if (maxr == nowl)
				{
					check = 1;
					break;
				}
				++ans;
			}
			if (check) printf("-1");
			else printf("%d", ans);
			printf("\n");
		}
	}
	return 0;
}

对于位置是p，半径是r的装置，根据勾股定理，可以算出 $d=\sqrt{r^2-(\frac{W}{2})^2}$，那么这个装置影响的范围是$[p-d,p+d]$，然后就和课内的摄像头问题的解法一样了。 需要注意忽略半径小于等于$\frac{W}{2}$的装置。

for (int i = 1; i <= n; i++) {
    cin >> p >> r;
    if (r <= W / 2) {
        n--;
        i--;
        continue;
    }
    double d = sqrt(r * r -  (W / 2) * (W / 2));
    t[i] = {p - d, p + d};
}