最短路 1006

Minimum Transport Cost

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 7   Accepted Submission(s) : 2

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

50 3 22 -1 43 0 5 -1 -122 5 0 9 20-1 -1 9 0 44 -1 20 4 05 17 8 3 11 33 52 4-1 -10

 

Sample Output

From 1 to 3 :Path: 1-->5-->4-->3Total cost : 21From 3 to 5 :Path: 3-->4-->5Total cost : 16From 2 to 4 :Path: 2-->1-->5-->4Total cost : 17题意：给定一个无向图，-1表示无法到达。另外经过一些城要收税。给定起点和终点。求最小花费和路径。
思路：floy。
#include<iostream>#include<stdio.h>#include<string.h>#define inf 999999999using namespace std;int map[1005][1005],cost[1005];int path[1005][1005];int n;void floy(){    for(int k=1; k<=n; k++)        for(int i=1; i<=n; i++)            for(int 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];//经过k点                    path[i][j]=path[i][k];                }                else if(map[i][j]==map[i][k]+map[k][j]+cost[k])                {                    if(path[i][j]>path[i][k])//比较选小的，题目要求选择小的输出                        path[i][j]=path[i][k];                }            }}int main(){    int st,end;    while(scanf("%d",&n),n)    {       for(int i=1;i<=n;i++)       for(int j=1;j<=n;j++)       {           path[i][j]=j;           cin>>map[i][j];           if(map[i][j]==-1)           map[i][j]=inf;       }       for(int i=1;i<=n;i++)       cin>>cost[i];       floy();       while(cin>>st>>end)       {           if(st==-1&&end==-1)break;           int s=st;           int e=end;           printf("From %d to %d :\n",s,e);           printf("Path: %d",s);           while(st!=end)//输出路径           {               printf("-->%d",path[st][end]);               st=path[st][end];           }           cout<<endl;           printf("Total cost : %d\n\n",map[s][e]);       }    }    return 0;}/*50 3 22 -1 43 0 5 -1 -122 5 0 9 20-1 -1 9 0 44 -1 20 4 05 17 8 3 11 33 52 4-1 -10*/