TACTQUER- editorial



Author: Tuấn Anh Trần Đặng
Tester: Kamil Dębowski
Editorialist: Tuấn Anh Trần Đặng




Graph, Dynamic Programming


Given a weighted vertex-catus of N vertices answer Q queries about the shortest path between a pair of arbitrary vertices u and v.


We will mainly work on the DFS tree of the catus eg. performing DFS from 1 and only take the edges that connect a vertex to the unvisited vertex during the process.

Sub-problem 1

Calculate the shortest path from 1 to every vertex u.


Let’s call this d[u]. We can get d by doing a dynamic programming in the DFS process. When we visit u there are two situations:

  1. u is not belong to any cycle: d[u] = d[parent[u]] + length of the edge from parent[u] to u.
  2. u is in a cycle C. In this case let top[C] is the vertex closest to the root in C. d[u] = d[top[C]] + shortest path from top[C] to u.

Note that there are exactly two ways to go between two vertices in the same cycle C so the shortest path from top[C] to u is just either the distance L from top[C] to u in the DFS tree or length of cycle C - L. top[C] should be pre calculated.

Sub-problem 2

Calculate the distance disTopDown(u, v) between u and one of it descendants v (in the DFS tree).


  • If u is not belongs to any cycle then disTopDown(u, v) = d[v] - d[u]. This is correct since when we go from 1 to v we must go to u first.
  • If u is belong to the cycle C. When go from 1 to v we may not go to u since we have two ways to pass through the cycle C. However we will have to go to the vertex b that is the vertex closest to v in C. We have that disTopDown(u, v) = d[v] - d[b] + shortest distance from b to u. Again since b and u are in the same cycle the distance between them can be easily calculated.

So it seems like our solution is quite obvious now: distance between two vertices u and v is disTopDown(lca(u, v), u) + disTopDown(lac(u, v), v) where lca(u, v) is the lowest common ancester of u and v in the DFS tree. But we still missing one more piece: how to find b. Let get to the final sub-problem.

Sub-problem 3

We will make the problem a bit more genernal. Given the cycle C and the vertex v find the first vertex of C in the path from v to the root of the DFS tree.


Hint: we can use the similar technique that we used in finding LCA.

Order the cycle in the way that if cycle A lies on the path from root to cycle B then cycle A will get the larger order. One way is to use the order of the DFS completion of the top vertex eg. if the DFS process at top[B] finished later then B got larger order.
Now you may already guessed the solution. We trying to go from v toward the top as far as possible making sure that we never enter the cycle C. Let order[C] is the order of cycle C we have to make sure that the maximum order of cycles that has a vertex in the path from v toward the root does not larger or equal to order[C].

Recall the problem of finding LCA, we have to prepare f[u][i] is the 2^i th parent of u. The formular is quite simple:

  • f[u][0] = parent[u]
  • f[u][i] = f[f[u][i - 1]][i - 1]

Apply the same technique let maxOrder[u][i] is the maximum label from u to its 2^ith parent:

  • maxOrder[u][0] = max(order[u], order[parent[u]]) where order[u] is the order of the cycle contain u or -1 if is does not belong to any cycle.
  • maxOrder[u][i] = max(maxOrder[u][i - 1], maxOrder[f[u][i - 1]][i - 1]);

With maxOrder calculated we try to jump from v toward the root and makesure that we never go a vertex with a order larger than or equal to order[C]. If b exist it will be the parent of the vertex we ended up at since we tried to go as close to C as possible but never enter it.


The solutions contains three part:

  1. Initial DFS to prepare infomation of the cycles eg. top vertex, label …
  2. Second DFS to calculate d[] and maxLabel[][].
  3. Answer the queries: (distance from u to v) = disTopDown(lca(u, v), u) + disTopDown(lca(u, v), v);

Complexity: O((N + M)logN)

Author’s/Tester’s Solutions:

Setter’s solution
Tester’s solution

1 Like

I was very disappointed solving that problem, it is very sad that in Codechef you are giving such well-known and messy problem. For example, “BZOJ 2125” is the exactly the same problem. I had also seen a lot of variations of such problem. I’d love to see much more fresh problems here.


my solution using dijkstra … output is ok… but why it shows sigsev … please anyne tell :slight_smile:
using namespace std;

#define max 200009
#define INF 300100

int d[max];

struct data

int city,dist;
data(int a,int b)
    city = a;
    dist = b;

bool operator < (const data& p) const

    return dist>p.dist;


int Dijkstra(int start,int node,int end)
int i,j,u,v;

for(j=0; j<=node; j++)
    d[j] = INF;



d[start] =0;

    data top = Q.top();

    u = top.city;

    if( u == end ) return ( d[end] );

    for(i=0; i<G[u].size(); i++)
        v = G[u][i];

            d[v] = d[u]+cost[u][i];



int main()
int node,edge,i,j,k,l,m,x,y,z,start,end,cas;


    for(i=0; i<edge; i++)




    while( cas-- )

    printf( "%d\n",Dijkstra(start,node,end) );



return 0;


Please move the problem to practice, the given link doesn’t work.