HDOJ Minimum Transport Cost

 

Minimum Transport Cost

http://acm.hdu.edu.cn/diy/contest_showproblem.php?cid=12497&pid=1008

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 1

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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)

 

这个题用dijkstra 算法不好解，纠结了一上午后来还得要用ployd算法，效率低了点。

代码如下：

 

#include<iostream>
using namespace std;
const int maxnum=200;
const int maxint=9999999; 
int map[maxnum][maxnum]; //图
int tax[maxnum];//税金
int path[maxnum][maxnum]; //保存路径，用来输出
int n; //城市数量
void floyd(){
     int i,j,k,t;
         for( i=1;i<=n;i++)
             for( j=1;j<=n;j++)
                  path[i][j]=j; //从i到j第一个经过的点的为j
         for( k=1;k<=n;k++)
             for( i=1;i<=n;i++)
                  for( j=1;j<=n;j++){
                       t=map[i][k]+map[k][j]+tax[k];
                      if(t<map[i][j]){
                         map[i][j]=t;
                         path[i][j]=path[i][k];
                      }
                      else if(t==map[i][j])  //注意字典序
                           if(path[i][j]>path[i][k])
                                path[i][j]=path[i][k];               
                  }
}
int main(){
    int i,j,k;
    while(scanf("%d",&n)!=EOF && n){
          for( i=1;i<=n;i++)
               for( j=1;j<=n;j++){
                       scanf("%d",&map[i][j]);
                       if(map[i][j]==-1)
                           map[i][j]=maxint; //如果不相连就置为无穷大
                   }
          for( i=1;i<=n;i++)
              scanf("%d",&tax[i]);  //税金
          floyd();
          int s,t;   //始点和终点
          while(scanf("%d%d",&s,&t)!=EOF && s!=-1 && t!=-1){
                 printf("From %d to %d :\n",s,t);
                 printf("Path: %d",s);
                 k=s;
                 while(k!=t){
                       printf("-->%d",path[k][t]);
                       k=path[k][t];
                 }
                 printf("\nTotal cost : %d\n\n",map[s][t]);
          }
    }
     return 0;  
}