Problem:
Vampire

Mr. C is a vampire. If he is exposed to the sunlight directly, he turns
into ash.
Nevertheless, last night, he attended to the meeting of Immortal and
Corpse Programmers
Circle, and he has to go home in the near dawn.
Fortunately, there are many tall buildings around Mr. C's home, and
while the sunlight is blocked by the buildings, he can move around
safely.
The top end of the sun has just reached the horizon now.
In how many seconds does Mr. C have to go into his safe coffin?

To simplify the problem, we represent the eastern dawn sky as a 2-dimensional
x-y plane, where the x axis is horizontal and the y axis
is vertical, and approximate building silhouettes by rectangles
and the sun by a circle on this plane.

The x axis represents the horizon.
We denote the time by t, and the current time is t=0.
The radius of the sun is r and
its center is at (0, -r) when the time t=0.
The sun moves on the x-y plane in a uniform linear motion at
a constant velocity of (0, 1) per second.

The sunlight is blocked if and only if the entire region of the sun
(including its edge) is included in the union of the silhouettes
(including their edges) and the region below the horizon (y ≤ 0).

Write a program that computes the time of the last moment when the sunlight is blocked.

The following figure shows the layout of silhouettes and the position of the sun at
the last moment when the sunlight is blocked, that corresponds to the
first dataset of Sample Input below. As this figure indicates, there are possibilities
that the last moment when the sunlight is blocked can be the time t=0.

The sunlight is blocked even when two silhouettes share parts of their edges.

The following figure shows the layout of silhouettes and the position of the sun at the
last moment when the sunlight is blocked, corresponding to the second
dataset of Sample Input.
In this dataset the radius of the sun is 2 and there are two silhouettes:
the one with height 4 is in -2 ≤ x ≤ 0, and
the other with height 3 is in 0 ≤ x ≤ 2.

Input

The input consists of multiple datasets.
The first line of a dataset contains two
integers r and n separated by a space.
r is the radius of the sun and n is the number of silhouettes
of the buildings.
(1 ≤ r ≤ 20, 0 ≤ n ≤ 20)

Each of following n lines contains three integers
xli, xri, hi (1 ≤ i ≤ n) separated by a space.

These three integers represent a silhouette rectangle of a building.
The silhouette rectangle is parallel to the horizon, and its left and right edges
are at x = xli and x = xri, its top
edge is at y = hi, and its bottom edge is on
the horizon.
(-20 ≤ xli < xri ≤ 20,
0 < hi ≤ 20)

The end of the input is indicated by a line containing two zeros separated by a space.

Note that these silhouettes may overlap one another.

Output

For each dataset, output a line containing the number indicating the time t of the last moment when the sunlight is blocked.
The value should not have an error greater than 0.001.
No extra characters should appear in the output.

Sample Input

2 3
-2 -1 3
0 1 3
2 3 3
2 2
-2 0 4
0 2 3
2 6
-3 3 1
-2 3 2
-1 3 3
0 3 4
1 3 5
2 3 6
2 6
-3 3 1
-3 2 2
-3 1 3
-3 0 4
-3 -1 5
-3 -2 6
0 0

Output for the Sample Input

0.0000
3.0000
2.2679
2.2679

Solution in C++11 (judge verdict: Wrong Answer)

#include <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <algorithm>
#include <cassert>
#include <iomanip>


using namespace std;


int main()
{


	int r, n;
	while (cin >> r >> n && r){

		int xl[20], xr[20], h[20];
		int sx = 21;
		int th = 0;
		int gx = -21;
		int t;

		for (int i = 0; i < n; i++){
			cin >> xl[i] >> xr[i] >> h[i];
			if (sx > xl[i] || (sx == xl[i] && th < h[i])){
				sx = xl[i];
				th = h[i];
				t = i;
			}
			if (gx < xr[i]){
				gx = xr[i];
			}
		}

		map<int, int> ma;

		for (int x = -20; x <= 20; x++)ma[x] = 0;
		for (int x = sx; x <= gx; x++){

			bool ef = false; int eh = -1;
			if (xr[t] == x){
				ef = true; eh = 0;
			}
			int mh = th; int tt = t;
			for (int i = 0; i < n; i++){
				if (xl[i] == x && mh < h[i]){
					mh = h[i]; tt = i;
				}
				if (x >= xl[i] && x < xr[i] && ef){
					eh = h[i]; tt = i;
				}
			}
			if (mh != th){
				t = tt; th = mh;
			}
			else
				if (ef){
					th = eh; t = tt;
				}
			ma[x] = th;
		}

		const double PI = 3.14159265359;
		double ans = 1e9;
		for (double x = -r; x <= r; x += 1e-5/2.0){

			double y = -double(r) + sqrt(double(r)*double(r) - x*x);
			int tx = (x >= 0 ? int(x) : int(x - (1.0 - 1e-7)));
			ans = min(ans, double(ma[tx]) - y);

		}

		cout << setprecision(30) << ans << endl;

	}

	return 0;
}