PROBLEM LINKDIFFICULTYeasy PREREQUISITESgraph theory, Dijkstra's algorithm, transforming graph into similar graph PROBLEMYou are given a undirected weighted graph consisting of $n$ vertices, and $m$ edges. Initially there is no edge between any pair of first $K$ vertices. We add the edge between each pair of first $K$ vertices of weight $X$ (i.e. the first $K$ vertices form a clique). Given a source vertex $s$, find out the shortest distance of each node from vertex $s$. Finding the shortest distance in a weighted graph from a fixed source vertex is a classical problem. There is a famous greedy algorithm called Dijkstra's algorithm which can solve this problem in $\mathcal{O}((V + E) log (V))$ time. If we directly apply the Dijkstra's algorithm in our problem, the number of vertices, i.e. $V = n$, and the number of edges, i.e. $E = m$ + $\binom{K}{2}$ (due to the clique). Hence, the time complexity of this algorithm will be $\mathcal{O}((m + K^2) \, log (n))$, which is enough to solve the first subtask, but won't be sufficient to solve the final subtask. Solution by constructing another graphOne of the ways of dealing with such situations is to think whether it is possible to convert the given graph into another graph that doesn't have $\mathcal{O}(K^2)$ edges of the clique added in it, and yet it preserves the pairwise distances of the vertices of the original graph. We can do this in the following way. Let us create a fake vertex. This fake vertex will be connected to the each of the first $K$ vertices. The weight of each edge will be $\frac{X}{2}$. So now, instead of the first $K$ vertices being connected in the form of clique are connected to the fake vertex in the form of a star topology where the central node of star is the fake vertex. // Add an image. Thus we have $K$ edges in the modified graph instead of $K^2$ edges of the original graph. The remaining edges in the modified graph will remain same. Notice that the distances in the modified graph are precisely the same as that of in the old graph. Here the edge connecting vertices $u$ and $v$ (where $u, v \leq K$) with weight $X$ directly has been replaced with the $u$ being connected with fake vertex with distance $\frac{X}{2}$, and from the fake vertex with $v$ with distance $\frac{X}{2}$. Apply Dijkstra's algorithm on this modified graph will yield a time complexity of $\mathcal{O}((m + K) log (n))$ which will give you full score. A small detail which can be useful here. As the number $\frac{X}{2}$ will not be an integer when $X$ is odd. So, instead of dealing with nonintegers (reals) in the Dijkstra's algorithm, you can multiply all the distances by $2$ to make all the distances even. Finally you divide the final distances by 2 to get the actual answer. In this way, you will never have to deal with real numbers. Solution by applying the usual Dijkstra with a little twistYou can apply the usual Dijkstra's algorithm. Let $d[i]$ denote the distance of vertex $i$ from source vertex $s$. When you are first time processing one of the vertices belonging to the clique, say vertex $u$, you consider the distances of all other vertices in the clique to be $d[u] + X$. Afterwards, whenever you encounter a vertex in clique, you need not to go over the clique edges to relax the distances of clique vertices. This algorithm will make sure that your time complexity is $\mathcal{O}((m + K) log (n))$, as instead of processing all the $K^2$ edges of clique, you are processing only $K$ of them. A solution by trusting your intuition and implement direct Dijkstra with a bit modification.One might intuitively think that the number of relaxations in the Dijkstra's algorithm might be $m + K$ instead of $m + K^2$. Proof of this intuition lies in the understanding of previous algorithm. You can directly implement this algorithm. You maintain a set (balanced binary search tree) of the clique vertices sorted by their distances in the increasing order. You can find the set of vertices to relax by making a search in the set. This way, you will only be relaxing the vertices only among the necessary vertices in the clique. See the editorialist's solution for a sample implementation of this approach. SETTER'S SOLUTIONTESTER'S SOLUTION
This question is marked "community wiki".
asked 15 Apr, 22:09

The naive(TLE) solution would be:
The above code clearly has a complexity O(k^2) which will not pass in 1s Improvisation by adding a fake vertex(explained) 1.Create an extra node i.e if n is 5, you need to create a graph of 6 vertices. Let's call this node as 'z' 2.Traverse i < 1 to k , for each i, make a connection to z with incoming weight to z as x(given in the problem) and outgoing weight from z to i as 0 The above two steps will create a starlike graph in which you can travel to a given node from any node within 1 to k with cost only as x and also reduce the k^2 factor in the complexity to k. The code with the fake vertex(z) would be:
Visualization of first test case: answered 17 Apr, 21:51

Can anyone please tell why I am getting TLE in the last two testcases ? link to my code answered 17 Apr, 16:59

can anyone please tell me why i am getting wrong ans in second task?? https://www.codechef.com/viewsolution/13274717 answered 17 Apr, 19:22

I solved this problem with a trick. Initially I didn't insert the edges of the clique in the edge list. While I am executing the Dijkstra part whenever I am getting a node from the heap which is less than or equal to K, I am doing the following step. Step : Inserting its adjacent edges which connects to the rest vertices in the clique in the edge list and then performing normal Dijkstra. And I am doing this operation only for this node and the next node which is less than or equal to K i.e. if I encounter node which is less than or equal to K but previously I had visited 2 nodes which were less than or equal to K then I am performing normal Dijkstra. this P.S I noticed this pattern of visiting the clique part twice after applying Dijkstra on many graphs with cliques. answered 17 Apr, 22:40
Your "trick" is what the intended solution is! But instead of 2, visiting just 1 node in the initial K nodes in sufficient. Your code with "cli < 2" changed to "cli < 1" also passes: https://www.codechef.com/viewsolution/13353308
(18 Apr, 03:29)

My logic/algo for the question is correct but couldn't get AC in contest. After contest saw the solution of one of my senior :this The logic/algo is same just data structures differ he used set,I used priority_queue in dijkstra and some minor differences.I changed my code too but couldn't get AC. But I am getting TLE for second subtask and WA for the first subtask. Can anyone spot a bug? Thanks! Referenced solution:here My Solution: here answered 17 Apr, 23:15

I approached the problem with the way: "Solution by applying the usual Dijkstra with a little twist" but unable to get AC ,can't figure out ,how do I do d[u]+X because: Note: I didn't connect 1 to k nodes with X. 1)Now do dijkstra(s) ,but since cliqued ones are not connected yet,so how do I connect these nodes so that I can get d[u]+X ? answered 18 Apr, 19:18

i approached the problem using simple dijkstra algoithm by connecting all adjacent pair in clique getting tle in second subtask which is fine but for first subtask it gives WA for second testcase ,first is AC .cant able to find bug code link>link text
link
This answer is marked "community wiki".
answered 18 Apr, 23:09

The approach I used was:
Intuitively, I was convinced this would work and came up with some kind of a proof in my mind. But it was giving wrong answer for some subtasks. I tried a lot of test cases and they all worked. Could someone tell me whats wrong with the logic? answered 19 Apr, 11:54
