Problem:
 Problem E: Water Tank Mr. Denjiro is a science teacher.

Today he has just received a specially ordered water tank that
will certainly be useful for his innovative experiments on water
flow.       Figure 1: The water tank    The size of the tank is 100cm (Width) * 50cm (Height) * 30cm (Depth)
 (see Figure 1).

For the experiments, he fits several partition boards inside of
the tank parallel to the sideboards.

The width of each board is equal to the depth of the tank, i.e. 30cm.
The height of each board is less than that of the tank, i.e. 50 cm,
and differs one another.

The boards are so thin that he can neglect the thickness in the experiments.       Figure 2: The front view of the tank    The front view of the tank is shown in Figure 2. There are several faucets above the tank and he turns them on at
    the beginning of his experiment.  The tank is initially empty.
    Your mission is to write a
    computer program to simulate water flow in the tank. Input The input consists of multiple data sets. D is the number of the data sets.  D  DataSet 1  DataSet 2 ...  DataSet D  The format of each data set 
( DataSet d , 1 <= d <= D ) 
is as follows.  N  B 1  H 1  B 2  H 2 ...  B N  H N  M  F 1  A 1  F 2  A 2 ...  F M  A M  L  P 1  T 1  P 2  T 2 ...  P L  T L  Each line in the data set contains one or two integers.  N is the number of the boards he sets in the tank . B i and H i are the x -position (cm) and the height (cm) 
of the i -th board, 
where 1 <= i <= N .  H i s differ from one another.  You may assume the
    following. 0 < N < 10 , 0 < B 1 < B 2 < ... < B N < 100 , 0 < H 1 < 50 ,
0 < H 2 < 50 ,
 ..., 0 < H N < 50.  M is the number of the faucets above the tank . F j and A j are 
the x -position (cm) and the amount of water flow (cm 3 /second)
of the j -th faucet ,
where 1 <= j <= M . There is no faucet just above any boards .
Namely, none of F j is equal to B i . You may assume the following . 0 < M <10 , 0 < F 1 < F 2 < ... 
< F M < 100 , 0 < A 1 < 100, 
0 < A 2 < 100, ...
0 < A M < 100.  L is the number of observation time and location. P k is the x -position (cm) of the k -th
observation point. T k is the k -th observation time in seconds from the beginning. None of P k is equal to B i . You may assume the following . 0 < L < 10 , 0 < P 1 < 100,
0 < P 2 < 100, ..., 
0 < P L < 100 , 0 < T 1 < 1000000, 
0 < T 2 < 1000000, 
... ,
0 < T L < 1000000. Output For each data set,
your program should output L lines each containing 
one real number which represents the height (cm) of the water level specified 
by the x -position P k at the time T k . Each answer may not have an error greater than 0.001.
As long as this condition is satisfied,
you may output any number of digits after the decimal point. After the water tank is filled to the brim,
the water level at any P k is equal to the
height of the tank, that is, 50 cm. Sample Input 2
5
15 40
35 20
50 45
70 30
80 10
3
20 3
60 2
65 2
6
40 4100
25 7500
10 18000
90 7000
25 15000
25 22000
5
15 40
35 20
50 45
70 30
80 10
2
60 4
75 1
3
60 6000
75 6000
85 6000 Output for the Sample Input 0.666667
21.4286
36.6667
11.1111
40.0
50.0
30.0
13.3333
13.3333

Solution (Python):
import sys
readline = sys.stdin.readline
write = sys.stdout.write

def solve():
    N = int(readline())
    W = [(0, 50)]
    for i in range(N):
        b, h = map(int, readline().split())
        W.append((b, h))
    W.append((100, 50))
    M = int(readline())
    S = [0]*(N+1)
    Q = []
    for i in range(M):
        f, a = map(int, readline().split())
        for j in range(N+1):
            if W[j][0] < f < W[j+1][0]:
                S[j] += a
                break
    L = int(readline())
    for i in range(L):
        p, t = map(int, readline().split())
        Q.append((t, i, p))
    Q.sort(reverse=1)
    T = [0]*(N+1)
    C = N+1
    tc = 0
    ans = [50] * L
    EPS = 1e-9
    while 1:
        k = -1; tm = 10**18
        for i in range(C):
            if S[i] == 0:
                continue
            b0, h0 = W[i]; b1, h1 = W[i+1]
            t = (min(h0, h1) * (b1 - b0) * 30 - T[i]) / S[i]
            if t < tm:
                tm = t
                k = i
        assert k != -1
        b0, h0 = W[k]; b1, h1 = W[k+1]
        dt = (min(h0, h1) * (b1 - b0) * 30 - T[k]) / S[k]
        while Q and tc <= Q[-1][0] < tc + dt:
            t, i, p = Q.pop()
            for j in range(C):
                ba, ha = W[j]; bb, hb = W[j+1]
                if ba < p < bb:
                    dt0 = t - tc
                    ans[i] = (S[j] * dt0 + T[j]) / ((bb - ba) * 30)
                    break
        for i in range(C):
            T[i] += S[i] * dt
        if C == 1:
            break
        if h0 < h1:
            if abs(T[k-1] - h0 * (b0 - W[k-1][0]) * 30) < EPS:
                assert S[k-1] == 0
                S[k-1] = S[k]
                T[k-1] += T[k]
                S.pop(k); T.pop(k); W.pop(k)
                C -= 1
            else:
                j = k-1
                while T[j] == W[j][1]:
                    j -= 1
                S[j] += S[k]
                S[k] = 0
        else:
            if abs(T[k+1] - h1 * (W[k+2][0] - b1) * 30) < EPS:
                assert S[k+1] == 0
                S[k+1] = S[k]
                T[k+1] += T[k]
                S.pop(k); T.pop(k); W.pop(k+1)
                C -= 1
            else:
                j = k+1
                while T[j] == W[j+1][1]:
                    j += 1
                S[j] += S[k]
                S[k] = 0
        tc += dt
    for i in range(L):
        write("%.16f\n" % ans[i])
D = int(readline())
for i in range(D):
    solve()