hdu 1385 Minimum Transport Cost

     

                                                     Minimum Transport Cost

                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                      Total Submission(s): 11411    Accepted Submission(s): 3192

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

50 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 Output

From 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
题目大意：
    给出一个矩阵，矩阵中每个数字代表i到j的花费，每通过一个点需要交纳费用，现在要从一点到另一点
    需要用的最小花费，并输出路径，当花费一样时，应按字典序较小的那个路径输出。
解题思路：
    用一个二维数组记录路径中途径各点的前驱，之后输出路径，应注意花费相同时间按字典序较小那个输出。
代码：
#include<stdio.h>#include<string.h>#include<stdlib.h>#define inf 99999999int a[1010][1010];int c[1010][1010];int q[1010];int n;int Min(int a,int b){return a<b?a:b;}void zdl(){int i,j,k,min,c1;for(i=1;i<=n;i++)  for(j=1;j<=n;j++)    c[i][j]=j;for(k=1;k<=n;k++){for(i=1;i<=n;i++){if(a[i][k]!=inf)for(j=1;j<=n;j++){int t=a[i][k]+a[k][j]+q[k];if(a[i][j]>t)  {  a[i][j]=t;  c[i][j]=c[i][k];   }if(a[i][j]==t)  c[i][j]=Min(c[i][j],c[i][k]);}}}}void print(int x,int y){int i=x;printf("From %d to %d :\n",x,y);printf("Path: %d",x);i=x;while(i!=y){printf("-->%d",c[i][y]);i=c[i][y];}printf("\n");printf("Total cost : %d\n\n",a[x][y]);}int main(){int i,j,k,m,x,y;while(scanf("%d",&n),n!=0){for(i=1;i<=n;i++) for(j=1;j<=n;j++)  {    scanf("%d",&a[i][j]);    if(a[i][j]==-1)      a[i][j]=inf;}for(i=1;i<=n;i++)  scanf("%d",&q[i]);  zdl();while(scanf("%d%d",&x,&y),x+y>0){print(x,y);}}return 0;}