运输问题表上作业法c语言

运输问题是一种经典的线性规划问题，可以使用线性规划算法来解决。下面是一个使用 C 语言实现运输问题表上作业法的示例代码：

#include <stdio.h>

#define MAX 100

void input(int a[MAX][MAX], int n, int m);
void output(int a[MAX][MAX], int n, int m);
void transport(int a[MAX][MAX], int n, int m, int supply[MAX], int demand[MAX]);

int main() {
    int a[MAX][MAX], n, m, supply[MAX], demand[MAX];

    printf("Enter the number of sources: ");
    scanf("%d", &n);
    printf("Enter the number of destinations: ");
    scanf("%d", &m);

    printf("Enter the transportation cost table:\n");
    input(a, n, m);

    printf("Enter the supply values:\n");
    for (int i = 0; i < n; i++) {
        scanf("%d", &supply[i]);
    }

    printf("Enter the demand values:\n");
    for (int j = 0; j < m; j++) {
        scanf("%d", &demand[j]);
    }

    printf("Transportation cost table:\n");
    output(a, n, m);

    transport(a, n, m, supply, demand);

    return 0;
}

void input(int a[MAX][MAX], int n, int m) {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            scanf("%d", &a[i][j]);
        }
    }
}

void output(int a[MAX][MAX], int n, int m) {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            printf("%d\t", a[i][j]);
        }
        printf("\n");
    }
}

void transport(int a[MAX][MAX], int n, int m, int supply[MAX], int demand[MAX]) {
    int x[MAX][MAX] = {0}, u[MAX] = {0}, v[MAX] = {0}, min, i, j, k;

    for (k = 1; k <= n + m - 1; k++) {
        min = 1000000000;

        for (i = 0; i < n; i++) {
            for (j = 0; j < m; j++) {
                if (x[i][j] == 0 && a[i][j] - u[i] - v[j] < min) {
                    min = a[i][j] - u[i] - v[j];
                    k = i;
                    l = j;
                }
            }
        }

        x[k][l] = supply[k] < demand[l] ? supply[k] : demand[l];
        supply[k] -= x[k][l];
        demand[l] -= x[k][l];

        for (i = 0; i < n; i++) {
            if (x[i][l] != 0) {
                u[i] += min;
            }
        }

        for (j = 0; j < m; j++) {
            if (x[k][j] != 0) {
                v[j] -= min;
            }
        }
    }

    printf("Transportation matrix:\n");
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            printf("%d\t", x[i][j]);
        }
        printf("\n");
    }
}

该代码实现了一个简单的运输问题表上作业法程序，通过输入运输花费表、供应量和需求量，输出最终的运输方案。