hdu 1385(Minimum Transport Cost) Floyd algorithm

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

 

 

Source

Asia 1996, Shanghai (Mainland China)

 

Analysis:

After read the problem,we can see the problem can solve by Floyd algorithm.The N cities and the road between them make up a graph.But the prolem is not only  calculate the path cost between the two cities.The transportation fee is also consist of the tax  if the cargo passing other cities.So the problem has become more difficult. 

The transprtation free include  two parts:(1) The cost of the path we can use the Floyed algorithm to solve it.We regard the path cost as the edge of the weight. (2)The tax problem we can add a new array intermediate[],and the array we record the intermediate city.

Specific code:

#include<iostream>
using namespace std;
#define max 0xfffffff
int path_cost[100][100];
int intermediate[100][100];
int tax[100];
int N;
void Floyd()
{
 int i,j,k;
 for(i=1;i<=N;i++)       
  for(j=1;j<=N;j++)
   intermediate[i][j]=j;
 for(i=1;i<=N;i++)    //*heart code of the Floyd
  for(j=1;j<=N;j++)
   for(k=1;k<=N;k++)
   {
    int mediate_cost=path_cost[j][i]+path_cost[i][k]+tax[i];
    if(path_cost[j][k]>mediate_cost)
    {
     path_cost[j][k]=mediate_cost;
     intermediate[j][k]=intermediate[j][i];
    }
    if(path_cost[j][k]==mediate_cost)                //* the lexically smallest one
    {
     if(intermediate[j][k]>intermediate[j][i])
      intermediate[j][k]=intermediate[j][i];
    }
   }
}
int main()
{
 int i,j;
 while(scanf("%d",&N)&&N)
 {
  int length;
  for(i=1;i<=N;i++)
   for(j=1;j<=N;j++)
   {
    cin>>length;
    if(length==-1)
     path_cost[i][j]=max;
    else
     path_cost[i][j]=length;
   }
  for(i=1;i<=N;i++)
   cin>>tax[i];
  Floyd();
  int a,b;
  while(scanf("%d %d",&a,&b)&&(a!=-1||b!=-1))
  {
   printf("From %d to %d :\n",a,b);
   printf("Path: %d",a);
   int t=a;
   while(t!=b)
   {
    printf("-->%d",intermediate[t][b]);
    t=intermediate[t][b];
   }
   printf("\n");
   printf("Total cost : %d\n\n",path_cost[a][b]);
  }
 }
 return 0;
}