Minimum Transport Cost Floyd 输出最短路

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. 

InputFirst is N, number of cities. N = 0 indicates the end of input. 

The data of path cost, city tax, source and destination cities are given in the input, which is of the form: 

a11 a12 ... a1N 
a21 a22 ... a2N 
............... 
aN1 aN2 ... aNN 
b1 b2 ... bN 

c d 
e f 
... 
g h 

where aij is the transport cost from city i to city j, aij = -1 indicates there is no direct path between city i and city j. bi represents the tax of passing through city i. And the cargo is to be delivered from city c to city d, city e to city f, ..., and g = h = -1. You must output the sequence of cities passed by and the total cost which is of the form: 
OutputFrom c to d : 
Path: c-->c1-->......-->ck-->d 
Total cost : ...... 
...... 

From e to f : 
Path: e-->e1-->..........-->ek-->f 
Total cost : ...... 

Note: if there are more minimal paths, output the lexically smallest one. Print a blank line after each test case. 

Sample Input

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

Sample Output

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 = 1; k <= n; k++)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; 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
", u);
        return;
    }
    int k = path[u][v];
    printf("%d-->", u);
    Print(k, v);
}
int main()
{
    while (scanf("%d", &n), n)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                scanf("%d", &g[i][j]);
                path[i][j] = j;
                if (g[i][j] == -1) g[i][j] = INF;
            }
        }
        for (int i = 1; i <= n; i++)
            scanf("%d", &v[i]);
        int f, t;
        Floyd();
        while (scanf("%d%d", &f, &t), f != -1 && t != -1)
        {
            printf("From %d to %d :
Path: ", f, t);
            Print(f, t);
            printf("Total cost : %d

", g[f][t]);
        }
    }
}