CHEFRRUN - Editorial

editorial
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(Note: the editorials are written in a hurry and without having solved the problems, so there may be mistakes. Feel free to correct them / inform me.)

PROBLEM LINK:

Practice
Contest

Author: Dmytro Berezin
Tester: Sergey Kulik
Editorialist: Xellos

DIFFICULTY:

SIMPLE

PREREQUISITES:

graphs

PROBLEM:

There are N boxes in a circle. From box i, you can only move A_i+1 times clockwise. Count the starting boxes which you’ll reach again during such a walk.

QUICK EXPLANATION:

You’re given a so-called functional graph; count the vertices which lie on cycles. Find all cycles by moving from each vertex over unvisited vertices, then move along each cycle to count them in linear time.

EXPLANATION:

Let’s number the boxes 0 through N-1. Now, we can say that the boxes are vertices of a directed graph with exactly one outgoing edge from i to (i+A_i+1)\%N. This graph has a well-known property: if we move from any vertex along those edges, we’ll eventually encounter exactly one cycle and stay on it.

So let’s start in some vertex v_s and move along edges. If we encounter a vertex that was visited before (for some earlier starting vertex), we already visited the cycle that follows, so there’s nothing to do for that v_s. Otherwise, we’ll have to encounter a cycle that wasn’t visited before - as soon as we visit a vertex that was visited before, but for the current v_s, we can conclude that this vertex v_c lies on that new cycle. Then, it’s sufficient to move along this cycle until we get back to v_c and count the vertices visited on it.

After doing that for all v_s, all cycles had to be visited, so we’ve counted the number of vertices on cycles - th answer to our problem. Time and memory complexity: O(N), since we can’t visit any vertex more than twice this way.

AUTHOR’S AND TESTER’S SOLUTIONS:

The author’s solution can be found here.
The tester’s solution can be found here.

RELATED PROBLEMS:

2

Hi I cannot view the solutions. I am getting access denied?

3

Hello ,access is denied to view the tester’s and author’s solution . Also if someone could see my solution and tell me where i went wrong : https://www.codechef.com/viewsolution/18711836

4

Since the editorial solutions are not available at present, my Python 3.5 solution is:

# https://www.codechef.com/problems/CHEFRRUN

for _ in range(int(input())):
    n = int(input())
    z = input().split()
    a = list(map(int, z[:n]))

    mag = []
    unused = [True]*n
    for ix in range(n):
        if unused[ix]:
            cyc = []
            c = ix
            while unused[c]:
                cyc.append(c)
                unused[c] = False
                c = (c + a[c] + 1)%n

            if c in cyc:
                mx = cyc.index(c)
                mag.extend(cyc[mx:])

    print(len(mag))
5

strongly connected components. kosaraju algorithm