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To solve this problem, we need to find the maximum number of 1s that can be placed in an n x m matrix such that no two 1s are in the same row or column, and 1s can only be placed in positions not marked as 0. This problem can be efficiently modeled and solved using the concept of maximum bipartite matching.

Approach

1. Problem Analysis: The problem constraints (no two 1s in the same row or column) are exactly the constraints of a bipartite matching problem. We can model the matrix as a bipartite graph where one set of nodes represents rows and the other set represents columns. An edge exists between row i and column j if the cell (i, j) is not marked as 0 (i.e., allowed to place a 1).

2. Key Insight: The maximum number of 1s we can place is the size of the maximum matching in this bipartite graph. Each edge in the matching corresponds to placing a 1 in the cell (i, j), and since it's a matching, no two edges share a row or column.

3. Algorithm: We use a Depth-First Search (DFS) based approach to find the maximum matching. For each row, we try to find an augmenting path to a column that is either unmatched or can be matched to another row.

Solution Code

def main():
    import sys
    input = sys.stdin.read().split()
    ptr = 0
    n = int(input[ptr])
    ptr += 1
    m = int(input[ptr])
    ptr += 1

    matrix = []
    for _ in range(n):
        row = list(map(int, input[ptr:ptr+m]))
        ptr += m
        matrix.append(row)

    # Build adjacency list: adj[i] is list of columns j where matrix[i][j] !=0
    adj = [[] for _ in range(n)]
    for i in range(n):
        for j in range(m):
            if matrix[i][j] != 0:
                adj[i].append(j)

    match_to = [-1] * m  # match_to[j] = row matched to column j

    def dfs(u, visited):
        for v in adj[u]:
            if not visited[v]:
                visited[v] = True
                if match_to[v] == -1 or dfs(match_to[v], visited):
                    match_to[v] = u
                    return True
        return False

    result = 0
    for u in range(n):
        visited = [False] * m
        if dfs(u, visited):
            result += 1

    print(result)

if __name__ == "__main__":
    main()

Explanation

1. Reading Input: The input is read and parsed into the matrix.
2. Adjacency List: We construct an adjacency list where each row points to the columns that are allowed to place a 1 (non-zero cells).
3. Matching Array: The match_to array keeps track of which row is matched to each column (initialized to -1 for unmatched columns).
4. DFS for Augmenting Path: For each row, we use DFS to find an augmenting path. If we find such a path, we update the matching array and increment the result.
5. Result: The size of the maximum matching is the maximum number of 1s we can place, which is printed as the output.

This approach efficiently finds the solution using standard bipartite matching techniques, ensuring correctness and optimal performance for typical problem constraints. The DFS-based method is straightforward to implement and works well for small to medium-sized matrices. For larger matrices, the Hopcroft-Karp algorithm can be used for faster computation, but the current solution is sufficient for most cases.

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