# Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2369    Accepted Submission(s): 584

Problem DescriptionThese 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 Input50 3 22 -1 43 0 5 -1 -122 5 0 9 20-1 -1 9 0 44 -1 20 4 05 17 8 3 11 33 52 4-1 -10

Sample OutputFrom 1 to 3 :Path: 1-->5-->4-->3Total cost : 21From 3 to 5 :Path: 3-->4-->5Total cost : 16From 2 to 4 :Path: 2-->1-->5-->4Total cost : 17

SourceAsia 1996, Shanghai (Mainland China)

RecommendEddyhttp://acm.hdu.edu.cn/showproblem.php?pid=1385  用Floyed算法，主要问题是记录路径，看了大牛做的才明白怎么样记录的。用map[a][b]记录从a到b最短路径的第一个经过的点！同时注意字典序排列代码：
`#include<stdio.h>#include<string.h>const int maxint=65533;#define MAXN 100int N;int distances[MAXN][MAXN];int map[MAXN][MAXN];//map[a][b]表示a到b的所有路径中的与a最近的一个int tax[MAXN];void Floyed(){  int i,j,k; for(i=1;i<=N;i++) for(j=1;j<=N;j++) map[i][j]=j; for( i=1;i<=N;i++) { for( j=1;j<=N;j++) for( k=1;k<=N;k++) { int t_dis=distances[j][i]+distances[i][k]+tax[i];  if(distances[j][k]>t_dis) { distances[j][k]=t_dis; map[j][k]=map[j][i];  } if(distances[j][k]==t_dis)//字典序  if(map[j][k]>map[j][i])map[j][k]=map[j][i];  }  } } int main(){ //freopen("test.in","r",stdin); //freopen("test.out","w",stdout); int i,j,A; while(scanf("%d",&N),N) { for(i=1;i<=N;i++) for(j=1;j<=N;j++) { scanf("%d",&A); if(A==-1)distances[i][j]=maxint; else distances[i][j]=A; } for(i=1;i<=N;i++)scanf("%d",&tax[i]); Floyed();  int a,b; while(scanf("%d%d",&a,&b)!=EOF) { if(a==-1&&b==-1)break; printf("From %d to %d :/n",a,b); printf("Path: %d",a); int t=a; while(t!=b) { printf("-->%d",map[t][b]); t=map[t][b]; }  //printf("-->%d/n",b); printf("/n"); printf("Total cost : %d/n/n",distances[a][b]);  }  } return 0; } `

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