HDU 1385 Minimum Transport Cost（Floyd+路径输出）

Minimum Transport Cost

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11367    Accepted Submission(s): 3176
[/b]

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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:

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

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预备知识:1.(Floyd)这里就不多说了2.Floyd路径输出:1.首先初始化对于所有的可联通的i j 两点path[i][j]=i；2.如果满足map[i][j]>map[i][k]+map[k][j],则pata[i][j]=path[i][k];3.遍历输出例：1-->2-->3 3-->4-->5dp[1][3]=dp[1][2]=2;dp[3][5]=dp[3][4]=4;d[1][5]=dp[1][3]=2;dp[1][5]=2;dp[2][5]=dp[2][3]=3;dp[3][5]=4;注意:本来以为直接遍历下来就是按照字典序排好的，但是发现wa所以仍然需要字典序的判断，原因的话因为1-->5-->4-->8-->7-->6-->3 和1-->2-->10-->3的话若不判断字典序k==7时就结束了不是输出1-->2-->10-->3而是1-->5-->4-->8-->7-->6-->3错误

#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define inf 0x00ffffff
int map[200][200];
int path[200][200];
int tax[200];
int main(int argc, char *argv[])
{
int n;
while(scanf("%d",&n),n)
{
memset(path,0,sizeof(path));
int i,j,k;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&map[i][j]);
if(map[i][j]!=-1)
path[i][j]=j;
else
map[i][j]=inf;
}
}
for(i=1;i<=n;i++)
scanf("%d",&tax[i]);
for(k=1;k<=n;k++)
{
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(map[i][j]>map[i][k]+map[k][j]+tax[k])
{
path[i][j]=path[i][k];
map[i][j]=map[i][k]+map[k][j]+tax[k];
}
else if(map[i][j]==(map[i][k]+map[k][j]+tax[k]))
{
if(path[i][j]>path[i][k])
{
path[i][j]=path[i][k];
map[i][j]=map[i][k]+map[k][j]+tax[k];
}
}
}
}
}
int start,end;
while(scanf("%d %d",&start,&end)!=EOF)
{
if(start==-1&&end==-1)
break;
printf("From %d to %d :\n",start,end);
printf("Path: %d",start);
int next=start;
while(next!=end)
{
printf("-->%d",path[next][end]);
next=path[next][end];
}
printf("\n",end);
printf("Total cost : %d\n\n",map[start][end]);
}
}
return 0;
}