指派问题 C语言代码

以下是一个简单的指派问题的 C 语言代码示例：

#include <stdio.h>

#define N 4

void assign(int cost_matrix[N][N], int assignment[N])
{
    int row_covered[N] = {0};
    int col_covered[N] = {0};
    int num_assigned = 0;
    int path_matrix[N][N] = {0};
    int path_count = 0;
    int row, col, i, j;

    // Step 1: Initialize the path matrix
    for (row = 0; row < N; row++) {
        for (col = 0; col < N; col++) {
            if (cost_matrix[row][col] == 0 && !row_covered[row] && !col_covered[col]) {
                path_matrix[row][col] = 1;
            }
        }
    }

    while (num_assigned < N) {
        // Step 2: Find an uncovered zero in the path matrix
        for (row = 0; row < N; row++) {
            for (col = 0; col < N; col++) {
                if (path_matrix[row][col] == 1 && !row_covered[row] && !col_covered[col]) {
                    // Step 3: Add the zero to the assignment
                    assignment[row] = col;
                    row_covered[row] = 1;
                    col_covered[col] = 1;
                    num_assigned++;
                }
            }
        }

        if (num_assigned == N) {
            break;
        }

        // Step 4: Find a non-covered row
        row = -1;
        for (i = 0; i < N; i++) {
            if (!row_covered[i]) {
                row = i;
                break;
            }
        }

        // Step 5: Initialize the path
        path_count = 0;
        for (j = 0; j < N; j++) {
            path_matrix[path_count][j] = 0;
        }

        // Step 6: Add the row to the path
        path_matrix[path_count][0] = row;
        col = assignment[row];
        path_matrix[path_count][1] = col;
        path_count++;

        // Step 7: Find the rest of the path
        while (1) {
            if (row_covered[row] == 0) {
                break;
            }

            for (i = 0; i < N; i++) {
                if (path_matrix[path_count - 1][i] == row) {
                    col = path_matrix[path_count - 1][i + 1];
                    break;
                }
            }

            row = -1;
            for (i = 0; i < N; i++) {
                if (cost_matrix[i][col] == 0 && !row_covered[i] && !col_covered[col]) {
                    row = i;
                    break;
                }
            }

            if (row == -1) {
                break;
            }

            path_matrix[path_count][0] = row;
            path_matrix[path_count][1] = col;
            path_count++;
        }

        // Step 8: Augment the path
        for (i = 0; i < path_count; i++) {
            if (row_covered[path_matrix[i][0]] == 1) {
                row_covered[path_matrix[i][0]] = 0;
                col_covered[path_matrix[i][1]] = 1;
                assignment[path_matrix[i][0]] = -1;
                assignment[path_matrix[i][1]] = path_matrix[i][0];
            } else {
                row_covered[path_matrix[i][0]] = 1;
                col_covered[path_matrix[i][1]] = 0;
                assignment[path_matrix[i][0]] = path_matrix[i][1];
                assignment[path_matrix[i][1]] = -1;
            }
        }

        // Step 9: Clear the covers and the path matrix
        for (i = 0; i < N; i++) {
            row_covered[i] = 0;
            col_covered[i] = 0;
            for (j = 0; j < N; j++) {
                path_matrix[i][j] = 0;
            }
        }
    }
}

int main()
{
    int cost_matrix[N][N] = {
        {3, 5, 1, 7},
        {2, 6, 4, 8},
        {9, 3, 5, 1},
        {7, 2, 8, 4}
    };
    int assignment[N] = {-1, -1, -1, -1};
    int i;

    assign(cost_matrix, assignment);

    printf("Assignment:\n");
    for (i = 0; i < N; i++) {
        printf("Worker %d assigned to task %d\n", i, assignment[i]);
    }

    return 0;
}

该示例实现了一个简单的指派问题算法，使用了匈牙利算法的变体。通过传入一个代价矩阵和一个指针数组，该函数可以计算出每个任务应该分配给哪个工人。在这个例子中，有4名工人和4个任务，每个工人都有不同的技能水平，每个任务需要不同的技能水平。该算法的目标是最小化总代价，即分配工人的总技能水平与任务所需技能水平之间的差异。