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.
//如果同时存在多条最短路径，按字典序选取最短路劲并且每个测试实例后输出一个空行
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
简单带路径保存的Floyd（新手没啥经验，不熟练，记下方便回味）
1 #include <iostream>
2 #include <cstdio>
3 using namespace std;
4 const int MAXINT = 1<<25;
5 int path[205][205];
6 int num[205][205];
7 int num1[205];
8 int n;
9 void floyd()
10 {
11 for(int k=1 ; k<=n ; k++)
12 for(int i=1 ; i<=n ; i++)
13 for(int j=1 ; j<=n ; j++)
14 {
15 if(num[i][j] > (num[i][k]+num[k][j]+num1[k]) )
16 {
17 num[i][j] = num[i][k]+num[k][j]+num1[k];
18 path[i][j] = path[i][k];
19 }
20 else if(num[i][j] == num[i][k]+num[k][j]+num1[k])
21 {
22 if(path[i][j] > path[i][k])
23 {
24 path[i][j] = path[i][k];
25 num[i][j] = num[i][k]+num[k][j]+num1[k];
26 }
27 }//attention 当存在多条最短路时，选取后驱小的路
28 }
29 }
30 void print(int a,int b)
31 {
32 if(a == b)
33 {
34 printf("%d\n",b);
35 return;
36 }
37 printf("%d-->",a);
38 a=path[a][b];
39 print(a,b);
40 }//路径输出函数
41
42 int main()
43 {
44 int i,j,a,b;
45 while(scanf("%d",&n),n)
46 {
47 for(i=1 ; i<=n ; i++)
48 for(j=1 ; j<=n ; j++)
49 {
50 scanf("%d",&num[i][j]);
51 if(num[i][j]==-1)
52 num[i][j]=MAXINT;
53 path[i][j]=j;//记后驱，方便输出
54 }
55 for(i=1 ; i<=n ; i++)
56 scanf("%d",&num1[i]);
57 floyd();
58 while(scanf("%d%d",&a,&b))
59 {
60 if(a == -1 && b == -1)
61 break;
62 printf("From %d to %d :\n",a,b);
63 printf("Path: ");
64 print(a,b);
65 printf("Total cost : %d\n\n",num[a][b]);
66 }
67 }
68 return 0;
69 }