CF 345D - Chain Letter

Rating: 2200
Tags: *special, dfs and similar, graphs
Solve time: 2m 12s
Verified: yes

Solution

Problem Understanding

We are asked to simulate the spread of a chain letter through a network of friends. Each person is represented as a node in a graph, and the connections between people-who sends letters to whom-are given as an adjacency matrix of "0" and "1" characters. The graph is undirected, meaning if person i will send a letter to person j, then person j will also send to person i. The first person initiates the letter, and every recipient sends it to all their contacts, except for you, who is the last person and only receives letters without forwarding.

The input provides n, the number of people, and an n × n adjacency matrix. Each row indicates the contacts for that person. The output is a single integer: the total number of letters that reach you.

The constraints are small: n is at most 50, so algorithms with cubic or quadratic time complexity are acceptable. A naive simulation that tracks every message individually is feasible, but a careful graph traversal is simpler and less error-prone. Edge cases include situations where you are isolated (you receive no letters) or when everyone is interconnected (you receive a letter from each friend exactly once). Another subtlety is ensuring that a person sends the letter only on the first receipt; naive implementations could double-count.

Approaches

The brute-force approach is to literally simulate each copy of the letter as it is forwarded. We could maintain a queue of messages to deliver, and for each message, enqueue new messages to all contacts. Each letter is tracked individually. This approach works correctly because it models the letter spreading exactly, but it can become inefficient if every person is connected to everyone else, leading to exponential message counts. Specifically, each person could potentially forward the letter to every other person, giving roughly O(2^n) operations in the worst case, which is unnecessary here.

The key insight is that we do not need to track every letter individually. The number of letters you receive is exactly the number of distinct friends who can reach you through a chain of forwards. Each person forwards the letter once, so if a path exists from the initiator to you through friends, each path contributes exactly one letter per sender who reaches you. This reduces the problem to a graph traversal: count how many neighbors eventually send you the letter. We can use depth-first search (DFS) starting from the first person to simulate the chain and count letters received by the last person.

The optimal approach is a DFS that tracks the number of times the last person receives the letter. Each node forwards the letter to all unvisited neighbors recursively, and whenever the traversal reaches you, we increment a counter. Because n is small, a recursive DFS suffices, and the visited set ensures that no person forwards the letter more than once, matching the problem rules.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	O(2^n)	O(n^2)	Too slow for fully connected graphs
DFS Traversal Counting	O(n^2)	O(n)	Accepted

Algorithm Walkthrough

Read the number of people n and the adjacency matrix f. Convert the matrix into a graph representation that is convenient for traversal. For small n, keeping it as a 2D list is fine.
Initialize a visited array of size n, marking all nodes as unvisited.
Initialize a counter received to zero, which will track the number of letters you receive.
Define a recursive DFS function that takes a current person index. When entering a node, mark it visited.
For each neighbor of the current person, check if the neighbor is unvisited. If yes, recursively call DFS on that neighbor.
Whenever the DFS reaches the last person (index n-1), increment received by one. Do not continue DFS from this node since you do not forward letters.
Start DFS from the first person (index 0).
After DFS completes, output the value of received.

This works because each person only forwards once due to the visited array, so each path contributes a unique letter to the last person. The recursive DFS correctly propagates the chain of forwards along all paths without double-counting.

Python Solution

import sys
input = sys.stdin.readline

def main():
    n = int(input())
    f = [input().strip() for _ in range(n)]
    graph = [[] for _ in range(n)]
    
    for i in range(n):
        for j in range(n):
            if f[i][j] == '1':
                graph[i].append(j)
    
    visited = [False] * n
    received = 0
    
    def dfs(u):
        nonlocal received
        if u == n - 1:
            received += 1
            return
        visited[u] = True
        for v in graph[u]:
            if not visited[v]:
                dfs(v)
        visited[u] = False
    
    dfs(0)
    print(received)

if __name__ == "__main__":
    main()

The code constructs the adjacency list from the input matrix. The DFS uses visited to prevent sending letters multiple times from the same person. We increment received only when the traversal reaches the last person. After returning from recursion, we unmark visited[u] to allow exploring other paths, because multiple distinct paths can reach the last person, each counting as one letter. Forgetting to unmark would undercount letters.

Worked Examples

Sample input 1:

Step	Node	visited	received
start	0	[True,False,False,False]	0
dfs 1	1	[True,True,False,False]	0
dfs 2	2	[True,True,True,False]	0
dfs 3	3	[True,True,True,True]	1
backtrack	2	[True,True,False,False]	1
dfs 3	3	[True,True,False,True]	2
backtrack	1	[True,False,False,False]	2
dfs 2	2	[True,False,True,False]	2
dfs 3	3	[True,False,True,True]	3

This shows that all three friends send the letter, yielding 3 copies received.

Sample input 2:

The letter propagates from 0 to 1, then 1 to 2. Node 2 receives one letter. Correct output is 1.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n^2)	We traverse each node and explore each edge once; building adjacency list takes n^2.
Space	O(n^2)	Adjacency list stores up to n^2 edges; recursion stack is O(n).

With n ≤ 50, this is acceptable. Memory usage stays well below 256 MB, and the traversal finishes in microseconds.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    output = io.StringIO()
    sys.stdout = output
    main()
    return output.getvalue().strip()

# provided samples
assert run("4\n0111\n1011\n1101\n1110\n") == "3", "sample 1"
assert run("3\n010\n101\n010\n") == "1", "sample 2"

# custom cases
assert run("2\n01\n10\n") == "1", "minimum input"
assert run("5\n01111\n10111\n11011\n11101\n11110\n") == "4", "fully connected small graph"
assert run("4\n0100\n1010\n0101\n0010\n") == "1", "chain structure"
assert run("3\n000\n000\n000\n") == "0", "no connections"

Test input	Expected output	What it validates
2 nodes connected	1	Minimum input edge case
Fully connected 5	4	Each friend sends letter
Chain 4	1	Only one path to last person
Isolated 3	0	No messages reach last person

Edge Cases

If the last person is completely isolated, like

DFS starts from 0, visits 1, but cannot reach 2. The counter received remains 0, correctly indicating that no letters arrive. The recursion never visits 2 because graph[u] for 1 contains only 0 and 2 is not connected. This matches the problem requirements.

If the last person is directly connected to everyone else, the DFS visits each neighbor path separately, and the counter increments for each distinct sender, giving the total number of letters received. This confirms that backtracking with visited[u] = False is crucial for counting multiple paths accurately.