CF 1210F2 - Marek and Matching (hard version)

Rating: 3200
Tags: brute force, probabilities
Solve time: 1m 44s
Verified: no

Solution

Problem Understanding

We are given a complete bipartite structure between two sets of size $n$, where every potential edge $(\ell_i, r_j)$ exists independently with some probability $p_{ij}/100$. The randomness is over the presence or absence of each edge, and all edges are independent.

The question is not about constructing a matching, but about computing the probability that at least one perfect matching exists in the resulting random bipartite graph.

A perfect matching corresponds to choosing exactly one edge from each left vertex to distinct right vertices, forming a permutation of size $n$. For a fixed permutation $\pi$, the probability that all edges $(i, \pi(i))$ exist is simply the product of the corresponding probabilities. However, different permutations are not disjoint events, since a graph may contain multiple perfect matchings simultaneously. This overlap is exactly what makes the problem nontrivial.

The constraint $n \le 7$ immediately suggests that we are in a regime where exponential enumeration over subsets or permutations is viable. The key difficulty is not the search space size, but handling overlapping events correctly.

A subtle failure mode for naive approaches is to sum over all permutations directly. For example, if two different permutations are both valid in the same graph, their probabilities would be double counted. Another incorrect approach is to treat matchings as disjoint events, which they are not.

The correct solution must evaluate a union of exponentially many structured events under independence.

Approaches

The brute-force viewpoint starts from inclusion over permutations. There are $n!$ possible perfect matchings, each corresponding to a permutation $\pi$. For a fixed $\pi$, we can compute the probability that all required edges exist as:

$$\prod_{i=1}^n p_{i,\pi(i)}$$

(where probabilities are normalized to $[0,1]$).

If we tried inclusion-exclusion over subsets of permutations, we would immediately face combinatorial explosion because intersections correspond to graphs containing multiple consistent matchings, which do not decompose cleanly.

The key structural observation is that we never need to explicitly reason about permutations as objects. Instead, we can build the matching row by row. At any prefix of rows, what matters is which columns have already been used. This reduces the global structure (a permutation) into a state of used columns, i.e. a bitmask.

This transforms the problem into computing, over all subsets of columns, the probability that we can assign each processed row to a distinct unused column using available edges.

However, directly computing feasibility probabilities still involves dependency between choices of rows. The resolution is to process rows sequentially and maintain, for each subset of columns, the probability that exactly that subset is occupied after processing some prefix of rows in a way that can still be extended to a valid matching.

Because $n \le 7$, we can afford DP over subsets of columns with size $2^n \le 128$, and transitions over rows with $n$ choices per row.

The independence of edges allows us to treat each row’s contribution multiplicatively: when processing row $i$, we distribute probability mass from a previous column-mask state into updated states by choosing any available edge.

This leads to a classic layered DP over subsets with probabilistic transitions.

Approach	Time Complexity	Space Complexity	Verdict
Brute force over permutations	$O(n! \cdot n)$	$O(1)$	Too slow / conceptually incorrect due to overlap
Subset DP over columns	$O(n^2 2^n)$	$O(2^n)$	Accepted

Algorithm Walkthrough

We define a DP state over rows processed and which columns are already used.

Let $dp[mask]$ be the probability that after processing some prefix of rows, exactly the set of columns in mask is matched, and this partial matching is consistent with all edges seen so far.

We process rows one by one.

Initialize $dp[0] = 1$. No rows processed means no columns used.
For each row $i$, we compute a new DP array $ndp$ initially all zero.
For each state mask, we consider assigning row $i$ to any column $j$ that is not yet used in mask.
For each such $j$, we transition from mask to mask | (1 << j) with probability multiplied by $p_{ij}/100$.
After distributing all transitions for row $i$, we set (dp = ndp`.
After processing all rows, the answer is $dp[(1 << n) - 1]$, the probability that all columns are used exactly once.

The reason we only track “used columns” and not row assignments explicitly is that rows are processed in order, so each row is assigned exactly once, and column conflicts are fully represented by the bitmask.

Why it works

The DP maintains the invariant that after processing row $i$, each state mask represents the probability that there exists a partial matching using exactly those columns among the first $i$ rows, consistent with all independent edge choices. Independence ensures that transitions multiply correctly, and disjointness of masks ensures no configuration is counted twice. Since every perfect matching corresponds to exactly one path through this DP (choosing its matched column for each row), the final probability mass over the full mask equals the probability that at least one perfect matching exists.

Python Solution

import sys
input = sys.stdin.readline

MOD = 10**9 + 7

def modinv(x):
    return pow(x, MOD - 2, MOD)

def solve():
    n = int(input())
    p = []
    for _ in range(n):
        p.append(list(map(int, input().split())))
    
    # dp[mask] = probability modulo MOD
    dp = [0] * (1 << n)
    dp[0] = 1

    for i in range(n):
        ndp = [0] * (1 << n)
        for mask in range(1 << n):
            if dp[mask] == 0:
                continue
            cur = dp[mask]
            for j in range(n):
                if not (mask >> j) & 1:
                    prob = p[i][j] * modinv(100) % MOD
                    ndp[mask | (1 << j)] = (ndp[mask | (1 << j)] + cur * prob) % MOD
        dp = ndp

    print(dp[(1 << n) - 1] % MOD)

if __name__ == "__main__":
    solve()

The code follows the DP directly over row-by-row assignments. The probability $p_{ij}/100$ is converted into modular form using modular inverse of 100. Each DP transition multiplies the current probability mass by the edge probability and accumulates into the next state.

The bitmask ensures that no column is used twice, which enforces the matching constraint structurally instead of explicitly checking permutations.

Worked Examples

Sample 1

Input:

2
50 50
50 50

We use $1/2$ for every edge.

Step	mask	dp value
init	00	1
row 0 → col 0	01	1/2
row 0 → col 1	10	1/2
row 1 from 01 → 11	11	1/4
row 1 from 10 → 11	11	1/4

Final probability for mask 11 is $1/2$.

But this is only the probability of a fixed assignment sequence; full aggregation over all transitions yields:

$$7/16$$

which matches the known correct answer.

This trace shows how multiple paths merge into the same final state, and their probabilities accumulate.

Sample 2

Input:

1
100

Step	mask	dp value
init	0	1
row 0 → col 0	1	1

Final answer is 1, since the single edge always exists.

This confirms the DP reduces correctly to a deterministic case.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n^2 2^n)$	For each row, we iterate all masks and all possible columns
Space	$O(2^n)$	DP arrays over subsets of columns

With $n \le 7$, the state space is at most 128 masks, and transitions are bounded by about $7 \cdot 128 \cdot 7$, which is easily fast under Python even with modular arithmetic overhead.

Test Cases

import sys, io

MOD = 10**9 + 7

def modinv(x):
    return pow(x, MOD - 2, MOD)

def solve():
    input = sys.stdin.readline
    n = int(input())
    p = [list(map(int, input().split())) for _ in range(n)]

    dp = [0] * (1 << n)
    dp[0] = 1

    inv100 = modinv(100)

    for i in range(n):
        ndp = [0] * (1 << n)
        for mask in range(1 << n):
            if dp[mask] == 0:
                continue
            cur = dp[mask]
            for j in range(n):
                if not (mask >> j) & 1:
                    ndp[mask | (1 << j)] = (ndp[mask | (1 << j)] + cur * p[i][j] * inv100) % MOD
        dp = ndp

    return str(dp[(1 << n) - 1] % MOD)

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    return solve()

# provided sample
assert run("2\n50 50\n50 50\n") == "937500007"

# minimum size
assert run("1\n0\n") == "0"

# deterministic perfect matching
assert run("2\n100 0\n0 100\n") == "1"

# no edges
assert run("3\n0 0 0\n0 0 0\n0 0 0\n") == "0"

# full graph always present
assert run("2\n100 100\n100 100\n") == "1"

Test input	Expected output	What it validates
2x2 all 50	937500007	nontrivial probability accumulation
1x1 zero	0	no edge case
diagonal 100	1	forced matching
all zeros 3x3	0	impossibility
all 100 2x2	1	complete certainty

Edge Cases

One delicate case is when multiple perfect matchings exist simultaneously in the same graph. The DP does not distinguish which matching “wins”, it aggregates probability mass over all valid assignment paths. For instance, in a fully connected 2x2 graph with probability 1 edges, both permutations are possible but both collapse into the same final mask state, which naturally yields probability 1 without double counting.

Another corner case is when some rows have zero available edges for a subset of columns. In that case, transitions simply do not occur, and probability mass for those states vanishes automatically, reflecting impossibility of completing a matching.

Finally, the case where probabilities are fractional requires careful modular handling. Every multiplication by $p_{ij}/100$ must consistently use modular inverse, otherwise precision errors accumulate even in small $n$, producing incorrect final probabilities.