DIRVS - Direct Visibility

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Building the GSM network is a very expensive and complex task. Moreover, after the Base Transceiver Stations (BTS) are built and working, we need to perform many various measurements to determine the state of the network, and propose effective improvements to be made.

The ACM technicians have a special equipment for measuring the strength of electro-magnetic fields, the transceivers' power and quality of the signal. This equipment is packed into a huge knapsack and the technician must move with it from one BTS to another. Unfortunately, the knapsack have not enough memory for storing all of the measured values. It has a small cache only, that can store values for several seconds. Then the values must be transmitted to the BTS by an infrared connection (IRDA). The IRDA needs direct visibility between the technician and the BTS.

Your task is to find the path between two neighbouring BTSes such that at least one of those BTSes is always visible.

Input

There is a single positive integer T on the first line of input (equal to about 500). It stands for the number of test cases to follow. Each test case consists of a town description. For simplicity, a town is modelled as a rectangular grid of P x Q square fields. Each field is exactly 1 metre wide. For each field, a non-negative integer Zi,j is given, representing the height of the terrain in that place, in metres. That means the town model is made of cubes, each of them being either solid or empty. There are no "half solid" cubes.

The first line of each test case contains two integer numbers P and Q, separated by a single space, 1 <= P,Q <= 200. Then there are P lines each containing Q integer numbers separated by a space. These numbers are Zi,j, where 1 <= i <= P, 1 <= j <= Q and 0 <= Zi,j <= 5000. After the terrain description, there are four numbers R1, C1, R2, C2 on the last line of each test case. These numbers represent position of two BTSes, 1 <= R1,R2 <= P, 1 <= C1,C2 <= Q. The first coordinate (R) determines the row of the town, the second coordinate determines the column.

The technician is moving in steps (steps stands for Standard Technician's Elementary Positional Shift). Each step is made between two neighbouring square fields. That means the step is always in North, South, West or East direction. It is not possible to move diagonally. The step between two fields A and B (step from A to B) is allowed only if the height of the terrain in the field B is not very different from the height in the field A. The technician can climb at most 1 metre up or descend at most 3 metres down in a single step.

At the end of each step, at least one of the two BTSes must be visible. However, there can be some point "in the middle of the step" where no BTS is visible. This is OK and the data is handled by the cache. The BTS is considered visible, if there is a direct visibility between the unit cube just above the terrain on the BTSes coordinates and the cube just above the terrain on the square field, where the technician is. Direct visibility between two cubes means that the line connecting the centres of the two cubes does not intersect any solid cube. However, the line can touch any number of solid cubes. In other words, consider both the BTS and the technician being points exactly half metre above the surface and in the centre of the appropriate square field.

Note that the IRDA beam can go between two cubes that touch each other by their edge, although there is no space between them. It is because such a beam touches both of these two cubes but does not intersect any of them. See the last test case of the sample input for an example of such a situation.

Output

You are to find the shortest possible path from BTS (R1, C1) to BTS (R2, C2), meeting the above criteria. All steps must be done between neighbouring fields, the terrain must not elevate or descend too much, and at the end of each step, at least one BTS must be visible.

For each test case, print one line containing the sentence The shortest path is M steps long., where M is the number of steps that must be made. If there is no such path, output the sentence Mission impossible!.

Example

Sample Input:

4
5 5
8 7 6 5 4
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
1 1 5 1
5 8
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
9 9 9 9 9 9 9 2
2 2 2 2 2 2 2 2
1 2 5 1
5 8
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
9 9 9 9 9 9 9 2
2 2 2 2 2 2 2 2
1 5 5 1
6 12
5 5 5 5 1 5 5 5 5 5 5 5
5 5 5 5 1 5 5 5 5 5 5 5
5 5 5 5 9 5 5 5 5 5 5 5
5 9 1 5 5 5 5 5 5 5 5 5
5 5 9 5 5 5 5 5 5 5 5 5
5 5 9 5 5 5 5 5 5 5 5 5
6 1 3 12

Sample Output:

The shortest path is 10 steps long.
Mission impossible!
The shortest path is 14 steps long.
The shortest path is 18 steps long.

hide comments
kauai68: 2020-05-21 17:23:31

A* with LoS works well

creeon: 2020-05-21 09:57:30

Looks good

MAYANK NARULA: 2019-12-31 05:51:05

I think Brute force won't work with simple recursion and visibility check for line of sight intercept discontinuity

helloworld1101: 2019-10-22 19:31:16

If BTS is considered visible, does this mean any cube unit on the path between BTS and technician has the height smaller or equal to the smaller height of BTS and technician?

jigar_g_c123: 2019-03-02 15:28:29

Please give explain this question in Brif please please

Last edit: 2019-03-02 15:28:56
kumar gaurav: 2015-10-06 10:53:34

can we use cache for only one movement?

Rafail Loizou: 2015-03-25 16:19:54

still can not figure out how to test visibility
do i need to make use of advanced computational geometry skills?

Mayank Raj: 2013-09-01 21:21:57

Dijkshtra getting timed out in java :'(
do i need to do a fib heap implementation???

ABHISHEK: 2013-05-26 21:04:27

how to check the condition if technician is directly visible from any of the two BTS?

tantrum: 2013-05-22 15:29:09

will bfs work for this??


Added by:adrian
Date:2004-05-08
Time limit:13s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: NODEJS PERL6 VB.NET
Resource:ACM Central European Programming Contest, Prague 2000