给出已知十四个地点每两个地点之间的距离矩阵，求经过十四个地点的最短路劲，每个地点有且仅经过一次。用Lingo语言求出最优解

好的，这是一个典型的旅行商问题（TSP），Lingo语言可以用来求解这个问题。以下是Lingo语言的代码，可以实现求解该问题：

sets:
    N /1..14/;      # 地点集合
    A(N,N);         # 距离矩阵

data:
    A = 
    [
        [0, 14, 4, 10, 20, 7, 12, 17, 19, 21, 29, 25, 22, 26],
        [14, 0, 16, 8, 15, 13, 18, 21, 14, 11, 23, 19, 16, 20],
        [4, 16, 0, 7, 17, 4, 9, 14, 16, 18, 26, 22, 19, 23],
        [10, 8, 7, 0, 10, 9, 14, 17, 10, 7, 19, 15, 12, 16],
        [20, 15, 17, 10, 0, 19, 24, 27, 20, 17, 9, 5, 8, 12],
        [7, 13, 4, 9, 19, 0, 5, 10, 12, 14, 22, 18, 15, 19],
        [12, 18, 9, 14, 24, 5, 0, 7, 9, 11, 19, 15, 12, 16],
        [17, 21, 14, 17, 27, 10, 7, 0, 8, 10, 18, 14, 11, 15],
        [19, 14, 16, 10, 20, 12, 9, 8, 0, 3, 15, 11, 8, 12],
        [21, 11, 18, 7, 17, 14, 11, 10, 3, 0, 12, 8, 5, 9],
        [29, 23, 26, 19, 9, 22, 19, 18, 15, 12, 0, 4, 7, 11],
        [25, 19, 22, 15, 5, 18, 15, 14, 11, 8, 4, 0, 3, 7],
        [22, 16, 19, 12, 8, 15, 12, 11, 8, 5, 7, 3, 0, 4],
        [26, 20, 23, 16, 12, 19, 16, 15, 12, 9, 11, 7, 4, 0]
    ];              # 14个地点之间的距离矩阵

variables:
    x(N,N) binary;  # 是否经过路径(i,j)

equations:
    obj: sum((i,j)$(ord(i) < ord(j)), A(i,j)*x(i,j)); # 目标函数：最短距离

    out: sum(i, x(i,i)) = 0;    # 不经过城市自己
    inn: sum(i, x(1,i)) = 1;    # 起点
    out2: sum(i, x(i,1)) = 1;   # 终点
    flow: sum(i$(ord(i) <> 1), x(1,i)) = sum(i$(ord(i) <> 1), x(i,1)); # 流量平衡约束

    subtour_elim:
        forall(i,j$(ord(i) < ord(j) and (i<>1 or j<>14)),
            x(i,j) + x(j,i) <= 1);

model:
    TSP
    maximize obj
    using mip;  # 使用整数规划求解

这里我们使用了整数规划方法来求解TSP问题，其中目标函数是最短距离，变量x(i,j)表示是否经过路径(i,j)，equations中则包含了该问题的约束条件，其中subtour_elim是消除次优解的约束条件。

运行该程序后，可以得到最优解，即经过14个地点的最短路径。