POJ 1385

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4487    Accepted Submission(s): 1114

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

 1 //转载 

 2 //路径存储的是前一个点的后继 

 3 #include <iostream>

 4 #include <stdio.h>

 5 #include <memory.h>

 6 #include <queue>

 7 using namespace std;

 8 const int INF = 0x7fffff;//并不是整形最大值，防止溢出 

 9 const int N = 10000;

10 int map[N][N], tax[N], path[N][N];

11 int n;

12 void init()

13 {

14     int i, j;

15     for(i = 1; i <= n; i++)

16         for(j = 1; j <= n; j++)

17             if(i == j) map[i][j] = 0;

18             else map[i][j] = INF;

19 }

20 void input()

21 {

22     int i, j, k;

23     for(i = 1; i <= n; i++)

24         for(j = 1; j <= n; j++)

25         {

26             scanf("%d", &k);

27             if(k != -1) map[i][j] = k;

28             path[i][j] = j;

29         }

30     for(i = 1; i <= n; i++)

31         scanf("%d", &tax[i]);

32 }

33 void floyd()

34 {

35     int i, j, k, len;

36     for(k = 1; k <= n; k++)

37     {

38         for(i = 1; i <= n; i++)

39         {

40             for(j = 1; j <= n; j++)

41             {

42                 len = map[i][k] + map[k][j] + tax[k];

43                 if(map[i][j] > len)

44                 {

45                     map[i][j] = len;

46                     path[i][j] = path[i][k];    //标记到该点的前一个点

47                 }

48                 else if(len == map[i][j])   //若距离相同

49                 {

50                     if(path[i][j] > path[i][k]) //判断是否为字典顺序

51                         path[i][j] = path[i][k];

52                 }

53             }

54         }

55     }

56 }

57 void output()

58 {

59     int i, j, k;

60     while(scanf("%d %d", &i, &j))

61     {

62         if(i == -1 && j == -1) break;

63         printf("From %d to %d :\n", i, j);

64         printf("Path: %d", i);

65         k = i;

66         while(k != j)   //输出路径从起点直至终点

67         {

68             printf("-->%d", path[k][j]);

69             k = path[k][j];

70         }

71         printf("\n");

72         printf("Total cost : %d\n\n", map[i][j]);

73     }

74 }

75 int main()

76 {

77     while(scanf("%d", &n), n)

78     {

79         init();

80         input();

81         floyd();

82         output();

83     }

84 

85     return 0;

86 }

87 /*

88 C语言，二维数组申请 

89 int **map;

90 *map = (int *)malloc(sizeof(int)*(city));

91 for(i=0;i<city;i++)

92     map[i] = (int *)malloc(sizeof(int)*city);

93 */