hdu1385(floyd算法+最短路径)

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5639    Accepted Submission(s): 1417

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
   
   
   
   

 

本题的意思是给出一个顶点和边都带权值的图，然后询问个点间的最短路径长度并输出最短路径

本题可以用floyd算法一次直接求出各点间的最短路径

或用dijkstra算法在每次询问后求得单源最短路径

由于习惯性以为一次性求出要好,所以我用的floyd算法，仔细一想由于询问次数未知用dijkstra算法可能要好些。

//这是大神的代码，粘贴别人的

#include <iostream>
#include <fstream>
using namespace std; 

int const Max=1001; 
int a[Max][Max];//存放边的权值
int b[Max];//存放顶点的权值
int nex[Max][Max]; //next[i][j]用来保存i-->j的最短路径中i的最优后（即最近，
int N; 

void floyd()
//flyod算法求个顶点间的最短路径长度并记录路径
{
	int i,j,k,fee; 
	
	for(i=1;i<=N;i++)
		for(j=1;j<=N;j++)
			nex[i][j]=j; 
		

		for(k=1;k<=N;k++)
		{
			for(i=1;i<=N;i++)
			{
				if(i==k||a[i][k]==-1)
					continue;
				for(j=1;j<=N;j++)
				{
					if(a[k][j]==-1||j==k)
						continue; 
					fee = a[i][k]+a[k][j]+b[k]; 
					if(a[i][j]==-1||a[i][j]>fee)
					{
						a[i][j]=fee; 
						nex[i][j]=nex[i][k]; 
					}
					//选择字典序小的路径
					else if(a[i][j]==fee)
					{
						if(nex[i][j]>nex[i][k])
							nex[i][j]=nex[i][k];
					}
				}
			}
		}
}

void path(int i, int j)
//递归输出最短路径
{
	if(j==nex[i][j])
	{
		printf("%d-->%d\n",i,j); 
	}
	else 
	{
		printf("%d-->",i); 
		path(nex[i][j],j);  
	}
}



int main()
{
	int i, j; 
	int A, B;
	while(cin>>N && N)
	{
		for(i=1; i<=N; i++)
			for(j=1; j<=N; j++)
				cin>>a[i][j];  
			for(i=1; i<=N; i++)
				cin>>b[i]; 
			
			floyd(); 	
			while(cin>>A>>B && (A!=-1 || B!=-1))
			{			
				printf("From %d to %d :\n", A, B); 	
				if(A == B)				
					printf("Path: %d\n", A);
				else
				{
					printf("Path: ");
					path(A, B);
				}
				printf("Total cost : %d\n\n", a[A][B]);
			}
	}
	return 0; 
}

 

//这是我的代码，实在不堪入目
#include<iostream>
#define MAX 1001
using namespace std;
int g[MAX][MAX];
int cost[MAX][MAX];
int path[MAX][MAX];
int way[MAX];
int city[MAX];
int n;

void floyd()
{
	int k,i,j;
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
		{
			cost[i][j]=g[i][j];
			path[i][j]=i;
		}
	}

	for(k=1;k<=n;k++)
	{
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				//打印路径！！！！！！！！！！！！！！！！！！！！！
				if(cost[i][j]!=-1&&cost[i][k]!=-1&&cost[k][j]!=-1&&cost[i][j]>cost[i][k]+cost[k][j])
				{
					cost[i][j]=cost[i][k]+cost[k][j]+city[k];
					path[i][j]=path[k][j];
				}
				else if(cost[i][j]==-1&&cost[i][k]!=-1&&cost[k][j]!=-1)
				{
					cost[i][j]=cost[i][k]+cost[k][j]+city[k];
					path[i][j]=path[k][j];
				}
			}			
		}
	}
}



int main()
{
	int i,s,e,j,k;
	while(cin>>n)
	{
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
				scanf("%d",&g[i][j]);			
			for(i=1;i<=n;i++)
				scanf("%d",&city[i]);
			floyd();
			while(cin>>s>>e&&!(s==-1&&e==-1))
			{
				k=0;
				way[k]=e;
				while(path[s][way[k]]!=s)
				{
					k++;
					way[k]=path[s][way[k-1]];
				}
				k++;
				way[k]=s;
				cout<<"From "<<s<<" to "<<e<<" :"<<endl;
				cout<<"Path: ";
				for(i=k;i>0;i--)
					cout<<way[i]<<"-->";
				cout<<way[i]<<endl;
				cout<<"Total cost : "<<cost[s][e]<<endl;
			}			
	}
	return 0;
}