These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivered from one city to another. The transportation fee consists of two parts: 
The cost of the transportation on the path between these cities, and

a certain tax which will be charged whenever any cargo passing through one city, except for the source and the destination cities.

You must write a program to find the route which has the minimum cost.

5

0 3 22 -1 4

3 0 5 -1 -1

22 5 0 9 20

-1 -1 9 0 4

4 -1 20 4 0

5 17 8 3 1

1 3

3 5

2 4

-1 -1

0

From 1 to 3 :

Path: 1-->5-->4-->3

Total cost : 21

From 3 to 5 :

Path: 3-->4-->5

Total cost : 16

From 2 to 4 :

Path: 2-->1-->5-->4

Total cost : 17

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<sstream>

#include<algorithm>

#include<queue>

#include<deque>

#include<iomanip>

#include<vector>

#include<cmath>

#include<map>

#include<stack>

#include<set>

#include<fstream>

#include<memory>

#include<list>

#include<string>

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

#define MAXN 103

#define L 31

#define INF 1000000009

#define eps 0.00000001

int g[MAXN][MAXN],path[MAXN][MAXN], n, v[MAXN];//path[i][j] 表示路径i->j上 i之后的第一个结点

void Floyd()

{

    for (int k = ; k <= n; k++)

    {

        for (int i = ; i <= n; i++)

        {

            for (int j = ; j <= n; j++)

            {

                if (g[i][j] > g[i][k] + g[k][j] + v[k])

                {

                    g[i][j] = g[i][k] + g[k][j] + v[k];

                    path[i][j] = path[i][k];//将路径结点缩小范围确定

                }

                else if (g[i][j] == g[i][k] + g[k][j] + v[k])

                {

                    if (path[i][j] > path[i][k])

                    {

                        path[i][j] = path[i][k];//比较前面的字典序

                    }

                }

            }

        }

    }

}

void Print(int u, int v)//递归 从i->j 可以分解为 i->path[i][j]->path[path[i][j]][j]->,,,,,j

{

    if (u == v)

    {

        printf("%d\n", u);

        return;

    }

    int k = path[u][v];

    printf("%d-->", u);

    Print(k, v);

}

int main()

{

    while (scanf("%d", &n), n)

    {

        for (int i = ; i <= n; i++)

        {

            for (int j = ; j <= n; j++)

            {

                scanf("%d", &g[i][j]);

                path[i][j] = j;

                if (g[i][j] == -) g[i][j] = INF;

            }

        }

        for (int i = ; i <= n; i++)

            scanf("%d", &v[i]);

        int f, t;

        Floyd();

        while (scanf("%d%d", &f, &t), f != - && t != -)

        {

            printf("From %d to %d :\nPath: ", f, t);

            Print(f, t);

            printf("Total cost : %d\n\n", g[f][t]);

        }

    }

}