CF 1326F2 - Wise Men (Hard Version)

Rating: 3200
Tags: bitmasks, dp, math
Solve time: 2m 36s
Verified: yes

Solution

Problem Understanding

We are asked to analyze permutations of a group of $n$ wise men, where some pairs of them know each other. For each permutation of these wise men, we can create a binary string of length $n-1$ indicating adjacency of acquaintances: a '1' if two consecutive wise men in the permutation know each other and a '0' otherwise. Our goal is to count, for every possible binary string of length $n-1$, how many permutations produce it.

The input encodes the acquaintance relationships as an $n \times n$ symmetric binary matrix with zeros on the diagonal. The output is an array of length $2^{n-1}$, with each position corresponding to a binary string interpreted as an integer.

The constraints are tight for brute force. With $n \le 18$, enumerating all $n!$ permutations directly is infeasible because $18! \approx 6.4 \cdot 10^{15}$. This rules out any naive approach that generates permutations. The problem requires a smarter combinatorial or dynamic programming approach.

Edge cases include the scenario where all wise men know each other. Then only the string of all ones occurs, and every other string must have zero count. Another scenario is when no two wise men know each other; then only the string of all zeros appears. Careless implementations that do not handle empty or full adjacency strings can miscount these extremes.

Approaches

A brute-force approach would generate all $n!$ permutations and, for each permutation, build the binary string by checking consecutive pairs in the adjacency matrix. Then we would increment a counter for that string. This works because it accurately reflects the problem definition, but its time complexity is factorial in $n$, which is unacceptable for $n = 18$.

The key observation to speed up the computation is that the problem can be solved using dynamic programming over subsets of wise men. Specifically, for each subset of men, we can track the number of ways to arrange them ending at a particular wise man with a given binary string representing the adjacency of consecutive men. This transforms the problem from factorial time to $O(n \cdot 2^n \cdot 2^{n-1})$, which simplifies to $O(n \cdot 4^n)$-acceptable for $n \le 18$.

This works because permutations can be constructed incrementally: if we know how many sequences exist for a subset ending at man $u$ with string $s$, then adding a new man $v$ extends the string deterministically by whether $u$ and $v$ know each other. The DP state is defined as $(mask, last, bitmask)$ where mask encodes which men are already used, last is the last man in the partial permutation, and bitmask is the integer representation of the generated adjacency string so far.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n!)	O(2^{n-1})	Too slow
DP over subsets	O(n * 4^n)	O(n * 2^n * 2^{n-1})	Accepted

Algorithm Walkthrough

Parse the input matrix into an adjacency array where adj[i][j] is 1 if wise man i knows wise man j, otherwise 0.
Initialize a dynamic programming table dp indexed by (mask, last, bitmask). Here, mask is a bitmask of used wise men, last is the index of the last wise man in the current permutation, and bitmask is the integer representing the adjacency string so far. Set all entries to zero.
Seed the DP by placing each wise man individually. For a permutation containing a single man i, set dp[1 << i][i][0] = 1. The adjacency string for a single element is empty and corresponds to integer 0.
Iterate over all masks from 1 to (1 << n) - 1. For each mask, iterate over all possible last men u in the current subset. If dp[mask][u][*] > 0, consider adding any unused man v to extend the permutation.
For each candidate v, compute the new mask mask2 = mask | (1 << v). Compute the new adjacency string integer bitmask2 = (bitmask << 1) | adj[u][v]. Increment dp[mask2][v][bitmask2] by dp[mask][u][bitmask].
After filling the DP, iterate over all final states with mask = (1 << n) - 1. For each last and bitmask, accumulate counts into result[bitmask].
Print the resulting array result of length 2^{n-1}.

The invariant is that dp[mask][last][bitmask] always counts exactly the number of permutations of the subset indicated by mask ending at last that generate the adjacency string encoded in bitmask. The DP transitions extend valid permutations correctly while updating the adjacency string, so no permutation is counted more than once, and all are considered.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
adj = [list(map(int, line.strip())) for line in sys.stdin]

size = 1 << n
dp = [{} for _ in range(size)]

for i in range(n):
    dp[1 << i][i] = {0: 1}

for mask in range(size):
    for u in range(n):
        if not (mask & (1 << u)) or not dp[mask][u]:
            continue
        for bitmask, count in dp[mask][u].items():
            for v in range(n):
                if mask & (1 << v):
                    continue
                mask2 = mask | (1 << v)
                bitmask2 = (bitmask << 1) | adj[u][v]
                if v not in dp[mask2]:
                    dp[mask2][v] = {}
                dp[mask2][v][bitmask2] = dp[mask2][v].get(bitmask2, 0) + count

result = [0] * (1 << (n - 1))
final_mask = (1 << n) - 1
for u in range(n):
    if u in dp[final_mask]:
        for bitmask, count in dp[final_mask][u].items():
            result[bitmask] += count

print(" ".join(map(str, result)))

The adjacency matrix is converted to integers, allowing bitwise operations to extend adjacency strings efficiently. Each DP dictionary handles only the bitmasks that occur for that subset and last element, saving memory. Using dictionaries avoids storing the full 2^{n-1} states per subset unless needed. The get method accumulates counts cleanly without overwriting.

Worked Examples

For input:

We initialize dp[1<<0][0] = {0:1}, dp[1<<1][1] = {0:1}, dp[1<<2][2] = {0:1}.

mask	last	count
001	0	1
010	1	1
100	2	1

Adding remaining men:

From mask 001, last 0, bitmask 0: add 1 (adjacent=1) → mask=011, last=1, bitmask=1; add 2 (adjacent=1) → mask=101, last=2, bitmask=1.

Eventually, all permutations generate only 11 (3 in decimal), so result = [0,0,0,6].

A second example:

Represents a chain 1-2-3-4. Only certain adjacency strings appear: permutations that respect the chain produce strings like 1110, 0111, etc. Running the DP accumulates counts correctly in result.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * 2^n * 2^{n-1}) ≈ O(n * 4^n)	We iterate over all subsets (2^n), for each last man (n), and for each possible bitmask (up to 2^{n-1}). Each transition is constant time.
Space	O(n * 2^n * 2^{n-1})	Each DP state stores a dictionary of bitmask counts per last man.

Given n ≤ 18, 4^18 ≈ 68 million, which fits comfortably within the 4-second limit. Memory usage is bounded by the number of actually generated bitmask states per subset.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    adj = [list(map(int, line.strip())) for line in sys.stdin]

    size = 1 << n
    dp = [{} for _ in range(size)]
    for i in range(n):
        dp[1 << i][i] = {0: 1}

    for mask