hdu-1385 Minimum Transport Cost(Dijkstra最短路径算法)

Minimum Transport Cost

Time Limit:2000MS     Memory Limit:65536KB

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

Dijkstra算法详解：http://blog.sina.com.cn/s/blog_4b9aefc20100zu8h.html

/*求在n个城市中从s城市到e城市的最小花费。i到j的花费由j城市的税和该路的花费组成，起点终点不算税。该题注意：1测试数据中有i->i的情况；2相同花费的路径按字典顺序小的优先输出；*/#include<iostream>#include<stdio.h>#include<vector>#include<string>#include<algorithm>#include<string.h>using namespace std;#define maxl 100int tax[maxl],map[maxl][maxl];//tax保存税费，map保存输入的路费int cost[maxl][maxl];//从i到j的总花费int pre[maxl];//保存到达该城市的上一个城市int dist[maxl];//用于dijkstra算法中保存i与已到达城市的最短距离int f[maxl][maxl];//邻接表储存与i直接相通的城市void output(int s,int e,int &sum)//因为pre[i]中保存的是上个一城市，{                                //所以要从终点找到起点，从起点开始输出路径if(s==e){printf("Path: %d",s);return;}sum+=cost[pre[e]][e];output(s,pre[e],sum);printf("-->%d",e);}bool cmp(const int &a,const int &b){return dist[a]<dist[b];}int main(){int i,j,m,n,s,e;while(scanf("%d",&n)&&n){for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&map[i][j]);for(i=1;i<=n;i++)scanf("%d",&tax[i]);while(scanf("%d%d",&s,&e)){if(s==-1&&e==-1) break;for(i=1;i<=n;i++){f[i][0]=0;for(j=1;j<=n;j++){if(i==j) continue;if(map[i][j]==-1) cost[i][j]=0x3f3f3f3f;else {cost[i][j]=map[i][j]+tax[j];f[i][++f[i][0]]=j;}}}vector<int> vec;//保存尚未到达的城市for(i=1;i<=n;i++){if(i==s) dist[i]=0;else {dist[i]=cost[s][i];vec.push_back(i);}pre[i]=s;}sort(vec.begin(),vec.end(),cmp);while(!vec.empty()){int k=vec[0];vec.erase(vec.begin());for(i=1;i<=f[k][0];i++){int z=f[k][i];if((dist[k]+cost[k][z])<dist[z]){//Dijkstra算法dist[z]=dist[k]+cost[k][z];pre[z]=k;}else if((dist[k]+cost[k][z])==dist[z])//花费相同路径，找出这两条路径进行比较{int kk=k,zz=pre[z];string a,b;a+=char(z+'0');b+=char(z+'0');while(kk!=s) a=char(kk+'0')+a,kk=pre[kk];while(zz!=s) b=char(zz+'0')+b,zz=pre[zz];if(a<b){dist[z]=dist[k]+cost[k][z];pre[z]=k;}}}sort(vec.begin(),vec.end(),cmp);}int w=s,sum=0;printf("From %d to %d :\n",s,e);output(s,e,sum);if(s!=e) sum-=tax[e];//i->i输出0printf("\n");printf("Total cost : %d\n\n",sum);}}return 0;}