Minimum Transport Cost(floyd+二维数组记录路径)

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8860    Accepted Submission(s): 2331

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

题解：坑了好久，首先就是城市的收费问题，首尾不加，只加中间；其次路径输出，竟然要用二维数组记录路径，一维为毛就不行，再者竟然要字典序输出。。。我类个去啊，逗了一比，长个记性，以后对于二维的数组就二维坐标记录路径，还有递归输出路径也要熟练。。。

代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 const int INF=0x3f3f3f3f;
 4 const int MAXN=1010;
 5 int map[MAXN][MAXN];
 6 int N,pre[MAXN][MAXN],cost[MAXN];
 7 void initial(){
 8     for(int i=1;i<=N;i++)
 9         for(int j=1;j<=N;j++){
10                 pre[i][j]=j;//初始化
11             if(i-j)map[i][j]=INF;
12                 else map[i][j]=0;
13         }
14 }
15 void floyd(){
16     for(int k=1;k<=N;k++)
17         for(int i=1;i<=N;i++)
18             for(int j=1;j<=N;j++)
19                 /*if(i==k||j==k)continue;
20             else*/ if(map[i][j]>map[i][k]+map[k][j]+cost[k]){
21             map[i][j]=map[i][k]+map[k][j]+cost[k];//就加中间
22             pre[i][j]=pre[i][k];
23             }
24             else if(map[i][j]==map[i][k]+map[k][j]+cost[k])
25                 if(pre[i][j]>pre[i][k])
26                 pre[i][j]=pre[i][k];//字典序输出。。。。
27 }
28 void print(int x,int y){
29     if(x==y){
30             printf("%d\n",y);
31         return ;
32     }
33     printf("%d-->",x);
34         print(pre[x][y],y);
35 }
36 int main(){
37     int a,b,c;
38     while(~scanf("%d",&N),N){
39         initial();
40         for(int i=1;i<=N;i++)
41             for(int j=1;j<=N;j++){
42                 scanf("%d",&a);
43                 if(a==-1)map[i][j]=INF;
44                 else map[i][j]=a;//这里错了我半天。。。。以前if（j>i）这样的就wa
45             }
46             for(int i=1;i<=N;i++)scanf("%d",&cost[i]);
47             floyd();
48             while(scanf("%d%d",&a,&b),a!=-1&&b!=-1){
49                 printf("From %d to %d :\n",a,b);
50                 printf("Path: ");
51                 print(a,b);
52                 printf("Total cost : %d\n",map[a][b]);
53                 puts("");
54             }
55     }
56     return 0;
57 }