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Bipartite matching - Grafos

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/**********************
 * Bipartite matching * O(m*n^2))
 **********************
 * Given a bipartite graph represented as an m-by-n matrix, where graph[i][j]
 * is true iff there is an edge from pigeon i to hole j, computes the maximum
 * number of pigeons that can find a hole (one per pigeon) and an optimal
 * assignment.
 *      Formally, this is a stripped down version of Ford-Fulkerson with
 * DFS used to find an augmenting path. DFS does the job quickly because
 * capacities can only be 0 or 1.
 *
 * PARAMETERS:
 *      - graph: adjacency matrix as above.
 * RETURNS:
 *      - an integer corresponding to the number of assignments
 *      - matchL[m]: for each pigeon, a hole or -1
 *      - matchR[n]: for each hole, a pigeon or -1
 * DETAILS:
 *      - graph[m][n], matchL[n], matchR[m] and seen[m] are global arrays
 *      - main() initializes matchL[] and matchR[] to all -1.
 *      - main() does a loop over all pigeons i and in each iteration
 *          - clears seen[] to all 0
 *          - calls bpm(i) and increments the maxflow counter
 *      - bpm(i) returns true iff pigeon i can be assigned a hole
 * FIELD TESTING:
 *      - UVa 10080, 10092, 670, 259
 **/

#include 

#define M 128
#define N 128

bool graph[M][N];
bool seen[N];
int matchL[M], matchR[N];
int n, m;

bool bpm(int u)
{
    for(int v = 0; v < n; v++) if(graph[u][v]) {
            if(seen[v]) continue;

            seen[v] = true;

            if(matchR[v] < 0 || bpm(matchR[v])) {
                matchL[u] = v;
                matchR[v] = u;
                return true;
            }
        }

    return false;
}

int main()
{
    // Read input and populate graph[][]
    // Set m, n
    memset(matchL, -1, sizeof(matchL));
    memset(matchR, -1, sizeof(matchR));
    int cnt = 0;

    for(int i = 0; i < m; i++) {
        memset(seen, 0, sizeof(seen));

        if(bpm(i)) cnt++;
    }

    // cnt contains the number of happy pigeons
    // matchL[i] contains the hole of pigeon i or -1 if pigeon i is unhappy
    // matchR[j] contains the pigeon in hole j or -1 if hole j is empty
    return 0;
}
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