Minimum Transport Cost

hese 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

#include

#include

#include

#include

using namespace std;

 

const int INF = 99999999;

const int N = 505;

 

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

int n;

 

void init()

{

    int i, j;

    for(i = 1; i <= n; i++)

        for(j = 1; j <= n;j++)

            if(i == j) map[i][j]= 0;

            else map[i][j] = INF;

}

 

void input()

{

    int i, j, k;

    for(i = 1; i <= n; i++)

        for(j = 1; j <= n;j++)

        {

            scanf("%d",&k);

            if(k != -1) map[i][j]= k;

            path[i][j] = j;

        }

    for(i = 1; i <= n; i++)

        scanf("%d",&tax[i]);

}

 

void floyd()

{

    int i, j, k, len;

    for(k = 1; k <= n; k++)

    {

        for(i = 1; i <= n;i++)

        {

            for(j = 1; j <= n;j++)

            {

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

                if(map[i][j] >len)

                {

                    map[i][j] =len;

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

                }

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

                {

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

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

                }

            }

        }

    }

}

 

void output()

{

    int i, j, k;

    while(scanf("%d%d", &i, &j))

    {

        if(i == -1 && j== -1) break;

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

        printf("Path:%d", i);

        k = i;

        while(k != j)   //输出路径从起点直至终点

        {

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

            k = path[k][j];

        }

        printf("\n");

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

    }

}

 

int main()

{

    while(scanf("%d",&n), n)

    {

        init();

        input();

        floyd();

        output();

    }

 

    return 0;

}

 

其中我们用 path 数组记录经过路径其实 path 的定义如下  path[i][j] = k 表示是最短路径 i-……j  和 j 的直接前驱  为 k 即是：i-->...............-->k ->j

举例子：

      如  1-> 5->4   4->3->6  此时 path[1][6] = 0 ； 0表示 1->6 不通  当我们引入节点 k = 4 此时有 1->5->4->3->6 显然有 paht[1][6] = 3= paht[4][6] = paht[k][6]

如是有 path[i][j] = path[k][j] 

对于 1->5 相邻边我们可以在初始化时候有 paht[1][5] = 1;

如是对于最短路径1->5->4->3->6 有 paht[1][6] = 3;paht[1][3]= 4; paht[1][4] = 5; paht[1][5] =1 如此逆推不难得到最短路径记录值