HDU1385 Minimum Transport Cost（floyd+标记路径）

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11029    Accepted Submission(s): 3062

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

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

以为理解了floyd，其实还远远不够。

对于本题，不止是标记路径，还要求输出最小字典序；

比如有这样三条边 

1->2   1

2->3   1

1->3   2

则有两种路径1->2->3和1->3

1->2->3的字典序较小，这里输出1->2->3;

对于输出路径，大致的原理就是一句话：最短路径的子路径也是最短路径

过程我就不分析了，想想还是很好理解的

#include <iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<map>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;const int inf=0x3f3f3f3f;int e[105][105],p[105][105],cost[105];int n;void init(){    for(int i=0;i<=n;i++)    {        for(int j=0;j<=n;j++)        {            if(i==j)            e[i][j]=0;            else            e[i][j]=inf;            p[i][j]=j;        }    }}void floyd(){    for(int k=1;k<=n;k++)    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(e[i][k]!=inf&&e[k][j]!=inf)                {                    int tmp=e[i][k]+e[k][j]+cost[k];                    if(e[i][j]>tmp)                {                    e[i][j]=tmp;                    p[i][j]=p[i][k];                }                else if(e[i][j]==tmp&&p[i][j]>p[i][k])                p[i][j]=p[i][k];                }            }        }    }}int main(){   while(~scanf("%d",&n)&&n)   {       init();       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;           }       }       for(int i=1;i<=n;i++)scanf("%d",&cost[i]);       floyd();       int st,ed;       while(~scanf("%d%d",&st,&ed))       {           if(st==-1&&ed==-1)break;           printf("From %d to %d :\n",st,ed);           printf("Path: %d",st);           int u=st;           while(u!=ed)           {               printf("-->%d",p[u][ed]);               u=p[u][ed];           }           puts("");           printf("Total cost : %d\n\n",e[st][ed]);       }   }   return 0;}