f+g+a>=15 g+a+b>=18 a+b+c>=25 b+c+d>=19 c+d+e>=31 d+e+f>=45 e+f+g>=47 minz=a+b+c+d+e+f+g的C语言代码

下面是一个实现上述线性规划问题的C语言代码，使用GLPK库求解：

#include <stdio.h>
#include <glpk.h>

int main()
{
    glp_prob *lp;
    int ia[1+1000], ja[1+1000];
    double ar[1+1000], z;
    int ret;
    
    // 创建一个线性规划问题
    lp = glp_create_prob();
    glp_set_prob_name(lp, "example");
    
    // 添加变量
    glp_add_cols(lp, 7);
    glp_set_col_name(lp, 1, "f");
    glp_set_col_kind(lp, 1, GLP_CV);
    glp_set_col_bnds(lp, 1, GLP_LO, 0.0, 0.0);
    glp_set_col_name(lp, 2, "g");
    glp_set_col_kind(lp, 2, GLP_CV);
    glp_set_col_bnds(lp, 2, GLP_LO, 0.0, 0.0);
    glp_set_col_name(lp, 3, "a");
    glp_set_col_kind(lp, 3, GLP_CV);
    glp_set_col_bnds(lp, 3, GLP_LO, 0.0, 0.0);
    glp_set_col_name(lp, 4, "b");
    glp_set_col_kind(lp, 4, GLP_CV);
    glp_set_col_bnds(lp, 4, GLP_LO, 0.0, 0.0);
    glp_set_col_name(lp, 5, "c");
    glp_set_col_kind(lp, 5, GLP_CV);
    glp_set_col_bnds(lp, 5, GLP_LO, 0.0, 0.0);
    glp_set_col_name(lp, 6, "d");
    glp_set_col_kind(lp, 6, GLP_CV);
    glp_set_col_bnds(lp, 6, GLP_LO, 0.0, 0.0);
    glp_set_col_name(lp, 7, "e");
    glp_set_col_kind(lp, 7, GLP_CV);
    glp_set_col_bnds(lp, 7, GLP_LO, 0.0, 0.0);
    
    // 添加约束
    glp_add_rows(lp, 6);
    glp_set_row_name(lp, 1, "c1");
    ia[1] = 1, ja[1] = 1, ar[1] = 1.0;
    ia[2] = 1, ja[2] = 2, ar[2] = 1.0;
    ia[3] = 1, ja[3] = 3, ar[3] = 1.0;
    glp_set_mat_row(lp, 1, 3, ia, ja, ar);
    glp_set_row_bnds(lp, 1, GLP_LO, 15.0, 0.0);
    glp_set_row_name(lp, 2, "c2");
    ia[1] = 1, ja[1] = 2, ar[1] = 1.0;
    ia[2] = 1, ja[2] = 3, ar[2] = 1.0;
    ia[3] = 1, ja[3] = 4, ar[3] = 1.0;
    glp_set_mat_row(lp, 2, 3, ia, ja, ar);
    glp_set_row_bnds(lp, 2, GLP_LO, 18.0, 0.0);
    glp_set_row_name(lp, 3, "c3");
    ia[1] = 1, ja[1] = 3, ar[1] = 1.0;
    ia[2] = 1, ja[2] = 4, ar[2] = 1.0;
    ia[3] = 1, ja[3] = 5, ar[3] = 1.0;
    glp_set_mat_row(lp, 3, 3, ia, ja, ar);
    glp_set_row_bnds(lp, 3, GLP_LO, 25.0, 0.0);
    glp_set_row_name(lp, 4, "c4");
    ia[1] = 1, ja[1] = 4, ar[1] = 1.0;
    ia[2] = 1, ja[2] = 5, ar[2] = 1.0;
    ia[3] = 1, ja[3] = 6, ar[3] = 1.0;
    glp_set_mat_row(lp, 4, 3, ia, ja, ar);
    glp_set_row_bnds(lp, 4, GLP_LO, 19.0, 0.0);
    glp_set_row_name(lp, 5, "c5");
    ia[1] = 1, ja[1] = 5, ar[1] = 1.0;
    ia[2] = 1, ja[2] = 6, ar[2] = 1.0;
    ia[3] = 1, ja[3] = 1, ar[3] = 1.0;
    glp_set_mat_row(lp, 5, 3, ia, ja, ar);
    glp_set_row_bnds(lp, 5, GLP_LO, 31.0, 0.0);
    glp_set_row_name(lp, 6, "c6");
    ia[1] = 1, ja[1] = 6, ar[1] = 1.0;
    ia[2] = 1, ja[2] = 7, ar[2] = 1.0;
    ia[3] = 1, ja[3] = 1, ar[3] = 1.0;
    glp_set_mat_row(lp, 6, 3, ia, ja, ar);
    glp_set_row_bnds(lp, 6, GLP_LO, 45.0, 0.0);
    
    // 设置目标函数
    ia[1] = 1, ja[1] = 3, ar[1] = 1.0;
    ia[2] = 1, ja[2] = 4, ar[2] = 1.0;
    ia[3] = 1, ja[3] = 5, ar[3] = 1.0;
    ia[4] = 1, ja[4] = 6, ar[4] = 1.0;
    ia[5] = 1, ja[5] = 7, ar[5] = 1.0;
    glp_set_obj_coef(lp, 3, 1.0);
    glp_set_obj_coef(lp, 4, 1.0);
    glp_set_obj_coef(lp, 5, 1.0);
    glp_set_obj_coef(lp, 6, 1.0);
    glp_set_obj_coef(lp, 7, 1.0);
    glp_set_obj_dir(lp, GLP_MIN);
    
    // 求解线性规划问题
    glp_simplex(lp, NULL);
    ret = glp_get_status(lp);
    
    // 获取结果
    if (ret == GLP_OPT)
    {
        z = glp_get_obj_val(lp);
        printf("optimal solution z = %g\n", z);
    }
    else
    {
        printf("no optimal solution\n");
    }
    
    // 释放内存
    glp_delete_prob(lp);
    
    return 0;
}

在这个例子中，我们首先创建一个线性规划问题，然后添加变量和约束，设置目标函数，调用glp_simplex函数求解线性规划问题，并获取结果。最后，我们释放内存并结束程序。

需要注意的是，这个线性规划问题是一个混合整数规划问题，但我们没有对变量进行整数约束。如果需要对变量进行整数约束，可以使用GLPK的整数规划求解器。