2013
12-09

Minimum Transport Cost

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.

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:

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.

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

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

/*

1.经过每个节点必须缴纳一点的税。

2.如果有多条最短路径，寻找字典序最小的路径。

PS：我一开始用dijstra

后来迫不得已，改成floyd了。。。。。。。。

*/

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int MAXN=1202;
const int INF=99999999;
int map[MAXN][MAXN],path[MAXN][MAXN],n,cost[MAXN],st,ed;
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(map[i][j]>map[i][k]+map[k][j]+cost[k])
{
map[i][j]=map[i][k]+map[k][j]+cost[k];
path[i][j]=path[i][k];
}
else if(map[i][j]==map[i][k]+map[k][j]+cost[k])//寻找字典序更小的路径
{
if(path[i][j]>path[i][k])
{
path[i][j]=path[i][k];
}
}
}
}
int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
int i,j,w;
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
path[i][j]=j;
scanf("%d",&map[i][j]);
if(map[i][j]==-1) map[i][j]=INF;
}
}
for(i=1; i<=n; i++)
scanf("%d",&cost[i]);
Floyd();
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,v=ed;
while (u!=v)
{
printf ("-->%d", path[u][v]);
u=path[u][v];
}
printf("\n");
printf("Total cost : %d\n\n",map[st][ed]);
}
}
return 0;
}

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2. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？