HDU 1385 Minimum Transport Cost (Dijstra 最短路)

Minimum Transport Cost

http://acm.hdu.edu.cn/showproblem.php?pid=1385

Problem Description

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.

 

Input

First 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:

 

Output

From 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

 

 

最后再提供两组测试数据：

Sample Input

4
0 2 3 9
2 0 1 5
3 1 0 3
9 5 3 1
0 0 0 0
1 4
-1 -1

4
0 2 3 9
2 0 1 5
3 1 0 3
9 5 3 1
1 2 3 4
1 4
-1 -1

Sample Output

From 1 to 4 :
Path: 1-->2-->3-->4
Total cost : 6

From 1 to 4 :
Path: 1-->2-->4
Total cost : 9

 

解题思路：将最短路保存起来，最短路相同，比较字典序，输出字典序最小的那个方案，此题难点也是按字典序输出！

具体方案见源代码！

解题代码：

 // File Name: Minimum Transport Cost 1385.cpp
 // Author: sheng
 // Created Time: 2013年07月18日 星期四 23时36分58秒

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 using namespace std;

 const int max_n = ;
 const int INF = 0x3fffffff;
 int map[max_n][max_n], cos[max_n];
 int vis[max_n], dis[max_n];
 int cun[max_n];
 int bg, ed, n;

 int out(int x, int y, int bg)
 {
     if (x != bg)
         x = out (cun[x], x, bg);
     printf("%d%s", x, x == ed ? "\n" : "-->");
     return y;
 }

 int sort(int j, int k)
 {
     int path_1[max_n];
     int path_2[max_n];
     int len1 = , len2 = ;
     path_1[len1 ++] = j;
     for (int i = k; ; i = cun[i])
     {
         path_1[len1 ++] = i;
         if (i == bg)
             break;
     }
     for (int i = j; ; i = cun[i])
     {
         path_2[len2 ++] = i;
         if (i == bg)
             break;
     }
     len1 --;
     len2 --;
     int len = len1 < len2 ? len1 : len2;
     for (int i = ; i <= len; i ++)
     {
         if (path_1[len1 - i] < path_2[len2 - i])
             return ;
         if (path_1[len1 - i] > path_2[len2 - i])
             return ;
     }
     if (len1 < len2)
         return ;
     return ;

 }

 int main ()
 {
     while (~scanf ("%d", &n), n)
     {
         for (int i = ; i <= n; i ++)
             for (int j = ; j <= n; j ++)
                 scanf ("%d", &map[i][j]);
         for (int i = ; i <= n; i ++)
             scanf ("%d", &cos[i]);
         int T = ;
         while (scanf ("%d%d", &bg, &ed) && bg != - && ed != -)
         {

             memset(vis, , sizeof (vis));
             for (int i = ; i <= n; i ++)
             {
                 if (map[bg][i] != - && bg != i)
                 {
                     dis[i] = map[bg][i] + cos[i];
                     cun[i] = bg;
                 }
                 else  dis[i] = INF;
             }
             dis[bg] = ;
             vis[bg] = ;
             for (int i = ; i <= n; i ++)
             {
                 int k;
                 int min = INF;
                 for (int j = ; j <= n; j++)
                 {
                     if (!vis[j] && min > dis[j])
                     {
                         min = dis[j];
                         k = j;
                     }
                 }
                 vis[k] = ;
                 for (int j = ; j <= n; j ++)
                     if (!vis[j] && map[k][j] != -)
                     {
                         if ( dis[j] > dis[k] + map[k][j] + cos[j])
                         {
                             cun[j] = k;
                             dis[j] = map[k][j] + dis[k] + cos[j];
                         }
                         else if (dis[j] == dis[k] + map[k][j] + cos[j])//花费相同时，寻找字典序最小的方案
                         {
                             if (sort(j, k))//比较字典序
                                 cun[j] = k;
                         }
                     }

             }
             printf ("From %d to %d :\n", bg, ed);
             printf ("Path: ");
             out (ed, , bg);//输出路径
             printf ("Total cost : %d\n\n", bg == ed ?  : dis[ed] - cos[ed]);
         }
     }
     return ;
 }