zoj 1456 Minimum Transport Cost （Floyd+路径记录）

Minimum Transport Cost

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

求两个城市间的最小费用，如果经过某个城镇还要交税，起始点和终点除外。

路径要求字典序最小，path[i][j]记录从i到j的第二个元素的编号。

#include"stdio.h"
#include"string.h"
#include"vector"
#include"queue"
#include"iostream"
#include"algorithm"
using namespace std;
#define N 1005
const int inf=1000000;
int n,s,t;
int g

,c
;
int path

;
int min(int a,int b)
{
return a<b?a:b;
}
void Floyd()
{
int i,j,k;
for(k=1;k<=n;k++)
{
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(c[k]+g[i][k]+g[k][j]<g[i][j])
{
g[i][j]=g[i][k]+g[k][j]+c[k];
path[i][j]=path[i][k];
}
else if(c[k]+g[i][k]+g[k][j]==g[i][j])
{
path[i][j]=min(path[i][j],path[i][k]);
}
}
}
}
}
int main()
{
int i,j;
while(scanf("%d",&n),n)
{
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
path[i][j]=j;
scanf("%d",&g[i][j]);
if(g[i][j]==-1)
g[i][j]=inf;
}
for(i=1;i<=n;i++)
scanf("%d",&c[i]);
while(scanf("%d%d",&s,&t),s!=-1||t!=-1)
{
Floyd();
printf("From %d to %d :\n",s,t);
printf("Path: %d",s);
int tmp=s;
while(tmp!=t)
{
tmp=path[tmp][t];
printf("-->%d",tmp);
}
printf("\nTotal cost : %d\n\n",g[s][t]);
}
}
return 0;
}