Minimum Transport Cost

最新推荐文章于 2023-04-06 07:30:00 发布

原创最新推荐文章于 2023-04-06 07:30:00 发布 · 160 阅读

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本文介绍了一种算法，用于在存在运输费和过城税的情况下，寻找从一个城市到另一个城市的最低成本路径。通过使用Floyd-Warshall算法并进行适当的修改，可以找到并输出具有最小总费用的路径。

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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：就是最短路，多了个输出路径罢了。

代码：

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f

int n,e[1005][1005],v[1005],path[1005][1005];

void print(int x,int y)
{
   int t;
   if(x==y){
    printf("%d\n",y);
    return ;
   }
   t=path[x][y];
   printf("%d-->",x);
   print(t,y);
}

int main()
{
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
        {
            scanf("%d",&e[i][j]);
            if(e[i][j]==-1) e[i][j]=INF;
            path[i][j]=j;  //记录每条边的后驱
        }
        for(int i=1;i<=n;i++)
            scanf("%d",&v[i]);
        for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++)
               for(int j=1;j<=n;j++)
        {
            if(e[i][j]>e[i][k]+e[k][j]+v[k])
            {
                e[i][j]=e[i][k]+e[k][j]+v[k]; //注意要加上税收
                path[i][j]=path[i][k];  //通过k点松弛了，注意更新数据
            }
            else if(e[i][j]==e[i][k]+e[k][j]+v[k]) //收费一样，要注意按字典序记录路径
            {
                if(path[i][j]>path[i][k])
						path[i][j]=path[i][k];
            }
        }
        int st,en;
        while(~scanf("%d %d",&st,&en))
        {
            if(st==-1&&en==-1) break;
            printf("From %d to %d :\n",st,en);
            printf("Path: ");
            print(st,en);
            printf("Total cost : %d\n\n",e[st][en]);
        }
    }
    return 0;
}