hdu1385Minimum Transport Cost(最短路變種)

 

思路：

輸出路徑的最短路變種問題。。這個題目在於多組詢問，那麼個人覺得用floyd更加穩妥一點。還有就是在每個城市都有過路費，所以在floyd的時候更改一下松弛條件即可。。那麼輸出路徑怎麼辦呢？？我采用的是輸出起點的後繼而不是終點的前驅。。因為我們關心的是路徑字典序最小，關心的是起點的後繼。。。那麼打印路徑的時候就直接從前向後打印，這個和dijkstra的打印路徑稍有不同。。。最短路的打印參見傳送門

題目：

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7538 Accepted Submission(s): 1935


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)
Recommend Eddy | We have carefully selected several similar problems for you: 1142 1217 2112 2722 1598

代碼：

#include
#include
#include
#include
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=50+10;
int dis[maxn][maxn],path[maxn][maxn],n,cost[maxn];
int u,st,en;

void floyd()
{
    for(int k=1;k<=n;k++)
      for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        int tmp=dis[i][k]+dis[k][j]+cost[k];
        if(tmppath[i][k]))
        {
            dis[i][j]=tmp;
            path[i][j]=path[i][k];
        }
    }
}

void read_Graph()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        scanf(%d,&u);
        if(u==-1)
            dis[i][j]=INF;
        else
        {
            dis[i][j]=u;
            path[i][j]=j;
        }
    }
    for(int i=1;i<=n;i++)
       scanf(%d,&cost[i]);
}

void solve()
{
    while(~scanf(%d%d,&st,&en))
    {
        if(st==-1&&en==-1)  break;
        printf(From %d to %d :
,st,en);
        printf(Path: %d,st);
        int Gery=st;
        while(Gery!=en)
        {
            printf(-->%d,path[Gery][en]);
            Gery=path[Gery][en];
        }
        printf(
Total cost : %d

,dis[st][en]);
    }
}

int main()
{
    while(~scanf(%d,&n),n)
    {
        read_Graph();
        floyd();
        solve();
    }
    return 0;
}