1006 最短路

最新推荐文章于 2020-08-14 10:54:53 发布

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本文介绍了一个寻找两个城市间最低运输成本路径的算法。该算法综合考虑了路径费用和经过城市的税费，并通过弗洛伊德算法求解。文章还提供了一段C++实现代码，展示了如何计算最小总成本及其对应的路径。

Minimum Transport Cost

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

Total Submission(s) : 22 Accepted Submission(s) : 8

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)

题意：任意两点的最短距离，要求按字典序输出路径其实就是进行松弛的时候加一个大小判断就行但是路径的存储还是不熟练~

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define MAXN 100
const int maxint=65533;
int N;
int distances[MAXN][MAXN];
int map[MAXN][MAXN];//map[a][b]表示a到b的所有路径中的与a最近的一个

int tax[MAXN];
void Floyed()
{
  
    int i,j,k;
    for(i=1;i<=N;i++)
     for(j=1;j<=N;j++)
        map[i][j]=j;
    for( i=1;i<=N;i++)
    {
        for( j=1;j<=N;j++)
          for( k=1;k<=N;k++)
          {
              int t_dis=distances[j][i]+distances[i][k]+tax[i];
              
              if(distances[j][k]>t_dis)
              {
                  distances[j][k]=t_dis;
                  map[j][k]=map[j][i];
                
              }
              if(distances[j][k]==t_dis)//字典序 
                if(map[j][k]>map[j][i])
			map[j][k]=map[j][i];    
          }    
    }    
} 
int main()
{
   
    int i,j,A;
    while(cin>>N)
    {
    	if(N==0) break;
        for(i=1;i<=N;i++)
          for(j=1;j<=N;j++)
          {
              cin>>A;
              if(A==-1)
			   distances[i][j]=maxint;
              else 
			   distances[i][j]=A;
          }
        for(i=1;i<=N;i++)
		 cin>>tax[i];
        Floyed();   
        int a,b;
        while(cin>>a>>b)
        {
            if(a==-1&&b==-1)break;
            cout<<"From "<<a<<" to "<<b<<" :"<<endl;
            cout<<"Path: "<<a;
            int t=a;
            while(t!=b)
            {
                cout<<"-->"<<map[t][b];
                t=map[t][b];
            }    
           
           cout<<endl;
            cout<<"Total cost : "<<distances[a][b]<<endl<<endl;
           
        }     
    }
    return 0;    
}