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POVisiting points, algorithm
 

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    <p>I'm learning for the exam from algorithms and data structures (that I have in a month) and I can't manage with finding efficient algorithm for this problem:</p> <blockquote> <p>We are given 1 &lt;= n &lt;= 5000 points on the line. Each point has different natural coordinate 0 &lt;= d &lt;= 10^6 and (not necessarily different) natural <strong>point-time</strong> 0 &lt;= t &lt;= 10^9 in seconds. We can start at any point and in each second change our current coordinate by +/-1. The problem is to visit all the points in such order that every point is visited before elapsing his point-time. Find the minimum total time (in seconds) of this trip, or say it's impossible.</p> </blockquote> <p>For example, 5 points given (coordinate, point-time):</p> <p>(1,3), (3,1), (5,6), (8,19), (10,15), it's possible, when we take a trip visiting coordinates: 3 -> 1 -> 5 -> 8 -> 10, we got minimum total time, which is equal to: 11.</p> <p>My first idea was to sort all the points lexicographically by: (point-time, coordinate), and then visit them in this order. Of course, when there are points between i-th point and (i+1)-th point, we can visit them before visiting (i+1)-th point. But unfortunately, there is no argument why such greedy approach should work, despite the fact that it would be difficult to implement. Maybe I'm trying to solve it too quickly? n is small so, O(n^2) should be ok, I suppose.</p> <p>I found other examples of the input, thinking maybe it will help me finding the solution. But now I only see that I have to find one permutation of all possible $n!$ permutations.</p> <p><strong>Input examples:</strong></p> <p>points (also given by coordinate, point time respectively): (0,4), (1,2), (4,5): surprisingly (I think) we have to visit them: 0 -> 1 -> 4, because any different order does not satisfy condition in one before last sentence in problem text.</p> <p>points: (0,7), (1,2), (2,1), (3, 4), (4,11), the only funny way is: 2 -> 1 -> 3 -> 0 -> 4, which takes us 10 seconds.</p> <p>Can anyone help?</p>
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    1. COisn't that equivalent to finding a minimum spanning tree on the graph where each point is a node and the vertices are the time it takes to move from one node to another ?
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    2. CO@AdrienPlisson: I don't understand your reduction, in here nodes have "life time", and it is also a single path - not a combination of different paths. Maybe I am missing your line of thought, care to elaborate?
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    3. COWell an greedy approach would be to sort the un-visited points according to `(t - CURRENT_T) - (x - CURRENT_X)` in ascending order. First one is greediest choice and if it goes negative then recursion breaks.
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