*HDU 1385 最短路 路径

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10183    Accepted Submission(s): 2791

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

 

 

Source

Asia 1996, Shanghai (Mainland China)

 

题意：

从一个城市到另一个城市有油费和每经过一个城市要缴纳过路费，起点和终点不用交，问从起点到终点最少的费用。并打印路径。

代码：

//floyd 用一个path数组记录路径。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int mp[1003][1003];
int fei[1003],path[1003][1003];
int main()
{
    int n,a,b,t;
    while(scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&mp[i][j]);
            if(mp[i][j]==-1)
            mp[i][j]=10000007;
            path[i][j]=j;      //初始化路径
        }
        for(int i=1;i<=n;i++)
        scanf("%d",&fei[i]);
        for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(mp[i][j]>mp[i][k]+mp[k][j]+fei[k])  //加上tax.
            {
                mp[i][j]=mp[i][k]+mp[k][j]+fei[k];
                path[i][j]=path[i][k];
            }
            else if(mp[i][j]==mp[i][k]+mp[k][j]+fei[k]&&path[i][j]>path[i][k])
            path[i][j]=path[i][k];
        }
        while(scanf("%d%d",&a,&b))
        {
            if(a<0&&b<0) break;
            printf("From %d to %d :\n",a,b);
            int x=a;
            printf("Path: %d",a);
            while(x!=b)
            {
                printf("-->%d",path[x][b]);
                x=path[x][b];
            }
            printf("\n");
            printf("Total cost : %d\n\n",mp[a][b]);
        }
    }
    return 0;
}