关于#c++#的问题：用代码块功能插入代码

用代码块功能插入代码，请勿粘贴截图

运行数据：
0 0 0
0 1 13
0 2 0
0 3 4
0 4 0
1 0 13
1 1 0
1 2 15
1 3 0
1 4 5
2 0 0
2 1 0
2 2 0
2 3 12
2 4 0
3 0 4
3 1 0
3 2 12
3 3 0
3 4 0
4 0 0
4 1 0
4 2 6
4 3 3
4 4 0

#include
using namespace std;
const int n = 5;
const int e = 25;
int sum1[5];
int sum2[5];
int sum3[5];
int j = 0;
int i = 0;
int Min;
int l;
#define max 20
class Graph
{
public:
    int arcs[n + 1][n + 1];
    int a[n + 1][n + 1];
    int path[n + 1][n + 1];
    void floyd(Graph& t, const int n);
};
void Graph::floyd(Graph& t, const int n)
{
    int i, j, w;
    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
        {
            if (i == j)
                t.arcs[i][j] = 0;
            else 
                t.arcs[i][j] = max;
            for (int k = 0; k < e; k++)
            {
                cin >> i >> j >> w;
                t.arcs[i][j] = w;
            }
            cout << endl;
        }
    for(int i=0;i0; j < n; j++)
        {
            t.a[i][j] = t.arcs[i][j];
            if((i != j) && (t.a[i][j] < max))
                t.path[i][j] = i;
            else 
                t.path[i][j] = 0;
        }
    for (int k = 0; k < n; k++)
    {
        for(i=0;i0;j1、输出邻接矩阵
    for (i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            cout << t.arcs[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
    //2、输出邻接表
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < n; j++)
        {
            if (i != j)
            {
                int next = t.path[i][j];
                cout << j;
                cout << "<--" << i << " " << t.a[i][j] << " ";
            }
        }
        cout << endl;
    }
    cout << endl;
    //3、输出最短路径的邻接矩阵
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < n; j++)
        {
            cout << t.a[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
    //4、输出点到所有点的路程的和
    for (i = 0; i < n; i++)
    {
        sum1[i] = 0;
        for (int j= 0;  j< n; j++)
        {
            if (i != j)
            {
                cout << t.a[i][j] << ":";
                int next = t.path[i][j];
                cout << j;
                while (next != i)
                {
                    cout << "<--" << next;
                    next = t.path[i][next];
                }
                cout << "<--" << i << endl;
                sum1[i] += t.a[i][j];
            }
        }
        cout << " 点" << i << " 到所有点的路程和为：  " << sum1[i] << endl;
    }
    //5、逆输出所有点到点路径的和
    for (i = 0; i < n; i++)
    {
        sum2[i] = 0;
        for (j = 0; j < n; j++)
        {
            if (i != j)
            {
                cout << t.a[j][i] << ":";
                int next = t.path[j][i];
                cout << i;
                while (next != j)
                {
                    cout << "<--" << next;
                    next = t.path[j][next];
                }
                cout << "<--" << j << endl;
            }
            sum2[i] += t.a[j][i];
        }
        cout << " 所有的点到" << i << " 路程和为：  " << sum2[i] << endl;
    }
    //6、输出所有点往返路程的和
    for (int o = 0; o < n; o++)
    {
        sum3[o] = sum1[o] + sum2[o];
        cout << " 点" << o << " 的往返路程的和为： " << sum3[o] << endl;
    }
    Min = sum3[0];
    //7、输出最短路程和医院建址
    for (int h = 0; h < n; h++)
    {
        if (sum3[h] <= Min)
        {
            Min = sum3[h];
            l = h;
        }
    }
    cout << " 比较可得在点" << l << " 那里建立医院" << Min << "  为最短路程" << endl;
}

int main()
{
    Graph t;
    t.floyd(t, n);
    return 0;
}

运行结果及报错内容

我想要达到的结果