Hdu1385minimum transport cost (short-circuit variant)

Link: HuangJing
Idea: The most short-circuit variant of the output path .. This question lies in multiple groups of inquiries, so I personally think it is more secure to use Floyd. In addition, there is a toll in every city, so you can change the Loose Condition in Floyd .. What about the output path ?? I use the successor of the output starting point instead of the precursor to the end point .. Because we care about the smallest Lexicographic Order of paths and the successor of the starting point... The print path is printed directly from the front to the back, which is slightly different from the print path of Dijkstra... For the printing of the shortest path, see the shipping question: 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 descriptionthese 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 shoshould be delivered from one city to another. the transportation extends 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, Please t for the source and the destination cities.

You must write a program to find the route which has the minimum cost.
 
Inputfirst 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:
 
Outputfrom 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

50 3 22 -1 43 0 5 -1 -122 5 0 9 20-1 -1 9 0 44 -1 20 4 05 17 8 3 11 33 52 4-1 -10

 
Sample output

From 1 to 3 :Path: 1-->5-->4-->3Total cost : 21From 3 to 5 :Path: 3-->4-->5Total cost : 16From 2 to 4 :Path: 2-->1-->5-->4Total cost : 17

 
Sourceasia 1996, Shanghai (Mainland China)
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Code:

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#define INF 0x3f3f3f3fusing 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(tmp<dis[i][j]||(tmp==dis[i][j]&&path[i][j]>path[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 :\n",st,en);        printf("Path: %d",st);        int Gery=st;        while(Gery!=en)        {            printf("-->%d",path[Gery][en]);            Gery=path[Gery][en];        }        printf("\nTotal cost : %d\n\n",dis[st][en]);    }}int main(){    while(~scanf("%d",&n),n)    {        read_Graph();        floyd();        solve();    }    return 0;}