Hdu1385 Minimum Transport Cost （disjtra）

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

Total Submission(s): 865    Accepted Submission(s): 211

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

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

 

 

Source

Asia 1996, Shanghai (Mainland China)

分析：dijkatra最短路记录路径就行。我本以为dijkstra算出来就是字典序因为那个（i=1...n）,但wa了后分析发现字典序小的路径有可能因为刚开始那段路比较长而被省略，因此需要在路径长度相同时加上字典序判断。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;const int INF=0x3f3f3f3f;const int MAXN=1005;int n,g[MAXN][MAXN],vis[MAXN],p[MAXN],dis[MAXN];void showpath(int x){if(x==-1)return ;showpath(p[x]);printf("%d-->",x);}int zid(int a,int b,int cur){int p1[MAXN],p2[MAXN],i=2,j=2;p1[0]=cur;p2[0]=cur;p1[1]=a;p2[1]=b;while(a!=-1){p1[i++]=p[a];a=p[a];}while(b!=-1){p2[j++]=p[b];b=p[b];}while(1){i--,j--;if(i<0)return p1[1];else if(j<0)return p2[1];if(p1[i]>p2[j])return p2[1];else if(p1[i]<p2[j])return p1[1];}}void dijk(int a,int b){int i,j,k;memset(vis,0,sizeof vis);for(i=1;i<=n;i++)dis[i]=g[a][i],p[i]=a;dis[a]=0;vis[a]=1;p[a]=-1;for(i=1;i<n;i++){int tmp=INF,mk;for(j=1;j<=n;j++)if(!vis[j]&&dis[j]<tmp)tmp=dis[mk=j];vis[mk]=1;for(j=1;j<=n;j++)if(!vis[j]&&dis[mk]+g[mk][j]+g[n+1][mk]<dis[j])dis[j]=dis[mk]+g[mk][j]+g[n+1][mk],p[j]=mk;else if(!vis[j]&&dis[mk]+g[mk][j]+g[n+1][mk]==dis[j])//路径冲突时选择字典序小的{p[j]=zid(p[j],mk,j);}}printf("From %d to %d :\nPath: ",a,b);showpath(p[b]);printf("%d\nTotal cost : %d\n\n",b,dis[b]);}int main(){int i,j,a,b;while(scanf("%d",&n)&&n){for(i=1;i<=n+1;i++)for(j=1;j<=n;j++){scanf("%d",&g[i][j]);if(g[i][j]==-1)g[i][j]=INF;}while(~scanf("%d%d",&a,&b)&&(a!=-1||b!=-1)){dijk(a,b);}}return 0;}