题目：
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
思路：这是最短路径的问题，不过要求出所走的路径是什么？
我用的是Floyd算法，因为除了所走的初末位置其他点都需要税款，所以此时的最短经费就是所要的运费和税款。然后就是路径了，首先用二维数组b[][]去存储一点到另一个点的路径，也就是当当前的e[i][j]需要更新时，从一点去另一个点也需要重新规划路径。
注意：如果有更多的最小路径，则输出点最小的路径，也就是当e[i][j]不需要更新时，该点与另一个点取较小的。
程序代码：
#include<stdio.h> int e[2000][2000],n,a[5000],b[2000][2000]; int inf=999999999; void floyd() { int k,i,j,ma; for(i=1;i<=n;i++) for(j=1;j<=n;j++) b[i][j]=j; for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) { if(e[i][k]<inf&&e[k][j]<inf) { ma=e[i][k]+e[k][j]+a[k]; if(e[i][j]>ma) { e[i][j]=ma; b[i][j]=b[i][k]; } else if(e[i][j]==ma) { if(b[i][j]>b[i][k]) b[i][j]=b[i][k]; } } } /* for(i=1;i<=n;i++) for(j=1;j<=n;j++) printf("*%d ",b[1][j]); printf("\n");*/ } int main() { int i,j,w,x,y,t1,t2; while(scanf("%d",&n),n!=0) { for(i=1;i<=n;i++) for(j=1;j<=n;j++) { scanf("%d",&w); if(w==-1) w=inf; e[i][j]=w; } for(i=1;i<=n;i++) scanf("%d",&a[i]); floyd(); while(scanf("%d%d",&x,&y)!=EOF) { if(x==-1&&y==-1) break; t1=x; t2=y; printf("From %d to %d :\n",x,y); printf("Path: %d",x); while(x!=y) { printf("-->%d",b[x][y]); x=b[x][y]; } printf("\n"); printf("Total cost : %d\n",e[t1][t2]); printf("\n"); } } return 0; }
