运输问题表上作业法c语言

运输问题可以使用表上作业法来解决，以下是用C语言实现表上作业法的示例代码：

#include <stdio.h>
#define MAX 100

void northWestCorner(int m, int n, int supply[MAX], int demand[MAX], int cost[MAX][MAX]);
void printSolution(int m, int n, int allocation[MAX][MAX], int cost[MAX][MAX]);

int main() {
    int m, n, i, j;
    int supply[MAX], demand[MAX], cost[MAX][MAX];

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

    printf("Enter the supply values:\n");
    for (i = 1; i <= m; i++) {
        scanf("%d", &supply[i]);
    }

    printf("Enter the demand values:\n");
    for (j = 1; j <= n; j++) {
        scanf("%d", &demand[j]);
    }

    printf("Enter the cost matrix:\n");
    for (i = 1; i <= m; i++) {
        for (j = 1; j <= n; j++) {
            scanf("%d", &cost[i][j]);
        }
    }

    northWestCorner(m, n, supply, demand, cost);

    return 0;
}

void northWestCorner(int m, int n, int supply[MAX], int demand[MAX], int cost[MAX][MAX]) {
    int i, j, allocation[MAX][MAX] = {0};
    int totalCost = 0;

    i = 1;
    j = 1;
    while (i <= m && j <= n) {
        if (supply[i] < demand[j]) {
            allocation[i][j] = supply[i];
            demand[j] -= supply[i];
            totalCost += allocation[i][j] * cost[i][j];
            i++;
        } else {
            allocation[i][j] = demand[j];
            supply[i] -= demand[j];
            totalCost += allocation[i][j] * cost[i][j];
            j++;
        }
    }

    printSolution(m, n, allocation, cost);
    printf("Total Cost: %d\n", totalCost);
}

void printSolution(int m, int n, int allocation[MAX][MAX], int cost[MAX][MAX]) {
    int i, j;

    printf("Allocation Matrix:\n");
    for (i = 1; i <= m; i++) {
        for (j = 1; j <= n; j++) {
            printf("%d\t", allocation[i][j]);
        }
        printf("\n");
    }

    printf("Cost Matrix:\n");
    for (i = 1; i <= m; i++) {
        for (j = 1; j <= n; j++) {
            printf("%d\t", cost[i][j]);
        }
        printf("\n");
    }
}

在这段代码中，我们首先输入了运输问题的基本数据，即源点数、汇点数、供应量、需求量和成本矩阵。然后使用表上作业法计算出最优解，并打印出分配矩阵和成本矩阵以及总成本。

这里使用了两个函数，`northWestCorner()` 和 `printSolution()`。`northWestCorner()` 函数实现了表上作业法，`printSolution()` 函数用于打印出分配矩阵和成本矩阵。