Floyd Warshall算法原理和实现|DP-16

2021年4月13日09:48:51 发表评论 1,037 次浏览

本文概述

Floyd Warshall算法用于解决所有对最短路径问题。问题是在给定的边缘加权有向图中找到每对顶点之间的最短距离。

例子：

Input:
       graph[][] = { {0, 5, INF, 10}, {INF, 0, 3, INF}, {INF, INF, 0, 1}, {INF, INF, INF, 0} }
which represents the following graph
             10
       (0)------->(3)
        |         /|\
      5 |          |
        |          | 1
       \|/     |
       (1)------->(2)
            3       
Note that the value of graph[i][j] is 0 if i is equal to j 
And graph[i][j] is INF (infinite) if there is no edge from vertex i to j.

Output:
Shortest distance matrix
      0      5      8      9
    INF      0      3      4
    INF    INF      0      1
    INF    INF    INF      0

Floyd Warshall算法

第一步, 我们初始化与输入图矩阵相同的解矩阵。然后, 我们通过将所有顶点视为中间顶点来更新解矩阵。这个想法是一个接一个地拾取所有顶点并更新所有最短路径, 其中包括将所拾取的顶点作为最短路径中的中间顶点。当我们选择顶点数k作为中间顶点时, 我们已经将顶点{0, 1, 2, .. k-1}视为中间顶点。对于源顶点和目标顶点的每对(i, j), 都有两种可能的情况。

1)k不是从i到j的最短路径的中间顶点。我们将dist [i] [j]的值保持不变。

2)k是从i到j的最短路径中的中间顶点。如果dist [i] [j]> dist [i] [k] + dist [k] [我们将dist [i] [j]的值更新为dist [i] [k] + dist [k] [j] j]

下图显示了所有对最短路径问题中的上述最佳子结构属性。

Floyd Warshall算法| DP-161

以下是Floyd Warshall算法的实现。

C ++

//C++ Program for Floyd Warshall Algorithm 
#include <bits/stdc++.h>
using namespace std;
  
//Number of vertices in the graph 
#define V 4 
  
/* Define Infinite as a large enough
value.This value will be used for 
vertices not connected to each other */
#define INF 99999 
  
//A function to print the solution matrix 
void printSolution( int dist[][V]); 
  
//Solves the all-pairs shortest path 
//problem using Floyd Warshall algorithm 
void floydWarshall ( int graph[][V]) 
{ 
     /* dist[][] will be the output matrix 
     that will finally have the shortest 
     distances between every pair of vertices */
     int dist[V][V], i, j, k; 
  
     /* Initialize the solution matrix same 
     as input graph matrix. Or we can say 
     the initial values of shortest distances
     are based on shortest paths considering 
     no intermediate vertex. */
     for (i = 0; i <V; i++) 
         for (j = 0; j <V; j++) 
             dist[i][j] = graph[i][j]; 
  
     /* Add all vertices one by one to 
     the set of intermediate vertices. 
     ---> Before start of an iteration, we have shortest distances between all 
     pairs of vertices such that the 
     shortest distances consider only the 
     vertices in set {0, 1, 2, .. k-1} as
     intermediate vertices. 
     ----> After the end of an iteration, vertex no. k is added to the set of 
     intermediate vertices and the set becomes {0, 1, 2, .. k} */
     for (k = 0; k <V; k++) 
     { 
         //Pick all vertices as source one by one 
         for (i = 0; i <V; i++) 
         { 
             //Pick all vertices as destination for the 
             //above picked source 
             for (j = 0; j <V; j++) 
             { 
                 //If vertex k is on the shortest path from 
                 //i to j, then update the value of dist[i][j] 
                 if (dist[i][k] + dist[k][j] <dist[i][j]) 
                     dist[i][j] = dist[i][k] + dist[k][j]; 
             } 
         } 
     } 
  
     //Print the shortest distance matrix 
     printSolution(dist); 
} 
  
/* A utility function to print solution */
void printSolution( int dist[][V]) 
{ 
     cout<<"The following matrix shows the shortest distances"
             " between every pair of vertices \n" ; 
     for ( int i = 0; i <V; i++) 
     { 
         for ( int j = 0; j <V; j++) 
         { 
             if (dist[i][j] == INF) 
                 cout<<"INF" <<"     " ; 
             else
                 cout<<dist[i][j]<<"     " ; 
         } 
         cout<<endl; 
     } 
} 
  
//Driver code 
int main() 
{ 
     /* Let us create the following weighted graph 
             10 
     (0)------->(3) 
         |     /|\ 
     5 |     | 
         |     | 1 
     \|/ | 
     (1)------->(2) 
             3     */
     int graph[V][V] = { {0, 5, INF, 10}, {INF, 0, 3, INF}, {INF, INF, 0, 1}, {INF, INF, INF, 0} 
                     }; 
  
     //Print the solution 
     floydWarshall(graph); 
     return 0; 
} 
  
//This code is contributed by rathbhupendra

C

//C Program for Floyd Warshall Algorithm
#include<stdio.h>
  
//Number of vertices in the graph
#define V 4
  
/* Define Infinite as a large enough value. This value will be used
   for vertices not connected to each other */
#define INF 99999
  
//A function to print the solution matrix
void printSolution( int dist[][V]);
  
//Solves the all-pairs shortest path problem using Floyd Warshall algorithm
void floydWarshall ( int graph[][V])
{
     /* dist[][] will be the output matrix that will finally have the shortest 
       distances between every pair of vertices */
     int dist[V][V], i, j, k;
  
     /* Initialize the solution matrix same as input graph matrix. Or 
        we can say the initial values of shortest distances are based
        on shortest paths considering no intermediate vertex. */
     for (i = 0; i <V; i++)
         for (j = 0; j <V; j++)
             dist[i][j] = graph[i][j];
  
     /* Add all vertices one by one to the set of intermediate vertices.
       ---> Before start of an iteration, we have shortest distances between all
       pairs of vertices such that the shortest distances consider only the
       vertices in set {0, 1, 2, .. k-1} as intermediate vertices.
       ----> After the end of an iteration, vertex no. k is added to the set of
       intermediate vertices and the set becomes {0, 1, 2, .. k} */
     for (k = 0; k <V; k++)
     {
         //Pick all vertices as source one by one
         for (i = 0; i <V; i++)
         {
             //Pick all vertices as destination for the
             //above picked source
             for (j = 0; j <V; j++)
             {
                 //If vertex k is on the shortest path from
                 //i to j, then update the value of dist[i][j]
                 if (dist[i][k] + dist[k][j] <dist[i][j])
                     dist[i][j] = dist[i][k] + dist[k][j];
             }
         }
     }
  
     //Print the shortest distance matrix
     printSolution(dist);
}
  
/* A utility function to print solution */
void printSolution( int dist[][V])
{
     printf ( "The following matrix shows the shortest distances"
             " between every pair of vertices \n" );
     for ( int i = 0; i <V; i++)
     {
         for ( int j = 0; j <V; j++)
         {
             if (dist[i][j] == INF)
                 printf ( "%7s" , "INF" );
             else
                 printf ( "%7d" , dist[i][j]);
         }
         printf ( "\n" );
     }
}
  
//driver program to test above function
int main()
{
     /* Let us create the following weighted graph
             10
        (0)------->(3)
         |         /|\
       5 |          |
         |          | 1
        \|/     |
        (1)------->(2)
             3           */
     int graph[V][V] = { {0, 5, INF, 10}, {INF, 0, 3, INF}, {INF, INF, 0, 1}, {INF, INF, INF, 0}
                       };
  
     //Print the solution
     floydWarshall(graph);
     return 0;
}

Java

//A Java program for Floyd Warshall All Pairs Shortest
//Path algorithm.
import java.util.*;
import java.lang.*;
import java.io.*;
  
  
class AllPairShortestPath
{
     final static int INF = 99999 , V = 4 ;
  
     void floydWarshall( int graph[][])
     {
         int dist[][] = new int [V][V];
         int i, j, k;
  
         /* Initialize the solution matrix same as input graph matrix.
            Or we can say the initial values of shortest distances
            are based on shortest paths considering no intermediate
            vertex. */
         for (i = 0 ; i <V; i++)
             for (j = 0 ; j <V; j++)
                 dist[i][j] = graph[i][j];
  
         /* Add all vertices one by one to the set of intermediate
            vertices.
           ---> Before start of an iteration, we have shortest
                distances between all pairs of vertices such that
                the shortest distances consider only the vertices in
                set {0, 1, 2, .. k-1} as intermediate vertices.
           ----> After the end of an iteration, vertex no. k is added
                 to the set of intermediate vertices and the set
                 becomes {0, 1, 2, .. k} */
         for (k = 0 ; k <V; k++)
         {
             //Pick all vertices as source one by one
             for (i = 0 ; i <V; i++)
             {
                 //Pick all vertices as destination for the
                 //above picked source
                 for (j = 0 ; j <V; j++)
                 {
                     //If vertex k is on the shortest path from
                     //i to j, then update the value of dist[i][j]
                     if (dist[i][k] + dist[k][j] <dist[i][j])
                         dist[i][j] = dist[i][k] + dist[k][j];
                 }
             }
         }
  
         //Print the shortest distance matrix
         printSolution(dist);
     }
  
     void printSolution( int dist[][])
     {
         System.out.println( "The following matrix shows the shortest " +
                          "distances between every pair of vertices" );
         for ( int i= 0 ; i<V; ++i)
         {
             for ( int j= 0 ; j<V; ++j)
             {
                 if (dist[i][j]==INF)
                     System.out.print( "INF " );
                 else
                     System.out.print(dist[i][j]+ "   " );
             }
             System.out.println();
         }
     }
  
     //Driver program to test above function
     public static void main (String[] args)
     {
         /* Let us create the following weighted graph
            10
         (0)------->(3)
         |         /|\
         5 |          |
         |          | 1
         \|/     |
         (1)------->(2)
            3           */
         int graph[][] = { { 0 , 5 , INF, 10 }, {INF, 0 , 3 , INF}, {INF, INF, 0 , 1 }, {INF, INF, INF, 0 }
                         };
         AllPairShortestPath a = new AllPairShortestPath();
  
         //Print the solution
         a.floydWarshall(graph);
     }
}
  
//Contributed by Aakash Hasija

python

# Python Program for Floyd Warshall Algorithm
  
# Number of vertices in the graph
V = 4 
  
# Define infinity as the large enough value. This value will be
# used for vertices not connected to each other
INF  = 99999
  
# Solves all pair shortest path via Floyd Warshall Algorithm
def floydWarshall(graph):
  
     """ dist[][] will be the output matrix that will finally
         have the shortest distances between every pair of vertices """
     """ initializing the solution matrix same as input graph matrix
     OR we can say that the initial values of shortest distances
     are based on shortest paths considering no 
     intermediate vertices """
     dist = map ( lambda i : map ( lambda j : j , i) , graph)
      
     """ Add all vertices one by one to the set of intermediate
      vertices.
      ---> Before start of an iteration, we have shortest distances
      between all pairs of vertices such that the shortest
      distances consider only the vertices in the set 
     {0, 1, 2, .. k-1} as intermediate vertices.
       ----> After the end of a iteration, vertex no. k is
      added to the set of intermediate vertices and the 
     set becomes {0, 1, 2, .. k}
     """
     for k in range (V):
  
         # pick all vertices as source one by one
         for i in range (V):
  
             # Pick all vertices as destination for the
             # above picked source
             for j in range (V):
  
                 # If vertex k is on the shortest path from 
                 # i to j, then update the value of dist[i][j]
                 dist[i][j] = min (dist[i][j] , dist[i][k] + dist[k][j]
                                 )
     printSolution(dist)
  
  
# A utility function to print the solution
def printSolution(dist):
     print "Following matrix shows the shortest distances\
  between every pair of vertices"
     for i in range (V):
         for j in range (V):
             if (dist[i][j] = = INF):
                 print "%7s" % ( "INF" ), else :
                 print "%7d\t" % (dist[i][j]), if j = = V - 1 :
                 print ""
  
  
  
# Driver program to test the above program
# Let us create the following weighted graph
"""
             10
        (0)------->(3)
         |         /|\
       5 |          |
         |          | 1
        \|/     |
        (1)------->(2)
             3           """
graph = [[ 0 , 5 , INF, 10 ], [INF, 0 , 3 , INF], [INF, INF, 0 , 1 ], [INF, INF, INF, 0 ]
         ]
# Print the solution
floydWarshall(graph);
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

//A C# program for Floyd Warshall All 
//Pairs Shortest Path algorithm.
  
using System;
  
public class AllPairShortestPath
{
     readonly static int INF = 99999, V = 4;
  
     void floydWarshall( int [, ] graph)
     {
         int [, ] dist = new int [V, V];
         int i, j, k;
  
         //Initialize the solution matrix 
         //same as input graph matrix
         //Or we can say the initial 
         //values of shortest distances
         //are based on shortest paths 
         //considering no intermediate
         //vertex
         for (i = 0; i <V; i++) {
             for (j = 0; j <V; j++) {
                 dist[i, j] = graph[i, j];
             }
         }
  
         /* Add all vertices one by one to 
         the set of intermediate vertices.
         ---> Before start of a iteration, we have shortest distances
              between all pairs of vertices
              such that the shortest distances
              consider only the vertices in
              set {0, 1, 2, .. k-1} as 
              intermediate vertices.
         ---> After the end of a iteration, vertex no. k is added
              to the set of intermediate
              vertices and the set
              becomes {0, 1, 2, .. k} */
         for (k = 0; k <V; k++)
         {
             //Pick all vertices as source
             //one by one
             for (i = 0; i <V; i++)
             {
                 //Pick all vertices as destination
                 //for the above picked source
                 for (j = 0; j <V; j++)
                 {
                     //If vertex k is on the shortest
                     //path from i to j, then update
                     //the value of dist[i][j]
                     if (dist[i, k] + dist[k, j] <dist[i, j]) 
                     {
                         dist[i, j] = dist[i, k] + dist[k, j];
                     }
                 }
             }
         }
  
         //Print the shortest distance matrix
         printSolution(dist);
     }
  
     void printSolution( int [, ] dist)
     {
         Console.WriteLine( "Following matrix shows the shortest " +
                         "distances between every pair of vertices" );
         for ( int i = 0; i <V; ++i)
         {
             for ( int j = 0; j <V; ++j)
             {
                 if (dist[i, j] == INF) {
                     Console.Write( "INF " );
                 } else {
                     Console.Write(dist[i, j] + " " );
                 }
             }
              
             Console.WriteLine();
         }
     }
  
     //Driver Code
     public static void Main( string [] args)
     {
         /* Let us create the following
            weighted graph
               10
         (0)------->(3)
         |         /|\
         5 |         |
         |         | 1
         \|/     |
         (1)------->(2)
              3             */
         int [, ] graph = { {0, 5, INF, 10}, {INF, 0, 3, INF}, {INF, INF, 0, 1}, {INF, INF, INF, 0}
                         };
          
         AllPairShortestPath a = new AllPairShortestPath();
  
         //Print the solution
         a.floydWarshall(graph);
     }
}
  
//This article is contributed by 
//Abdul Mateen Mohammed

的PHP

<?php
//PHP Program for Floyd Warshall Algorithm 
  
//Solves the all-pairs shortest path problem
//using Floyd Warshall algorithm 
function floydWarshall ( $graph , $V , $INF ) 
{ 
     /* dist[][] will be the output matrix 
     that will finally have the shortest 
     distances between every pair of vertices */
     $dist = array ( array (0, 0, 0, 0), array (0, 0, 0, 0), array (0, 0, 0, 0), array (0, 0, 0, 0));
  
     /* Initialize the solution matrix same 
     as input graph matrix. Or we can say the 
     initial values of shortest distances are 
     based on shortest paths considering no 
     intermediate vertex. */
     for ( $i = 0; $i <$V ; $i ++) 
         for ( $j = 0; $j <$V ; $j ++) 
             $dist [ $i ][ $j ] = $graph [ $i ][ $j ]; 
  
     /* Add all vertices one by one to the set 
     of intermediate vertices. 
     ---> Before start of an iteration, we have 
     shortest distances between all pairs of 
     vertices such that the shortest distances 
     consider only the vertices in set 
     {0, 1, 2, .. k-1} as intermediate vertices. 
     ----> After the end of an iteration, vertex 
     no. k is added to the set of intermediate
     vertices and the set becomes {0, 1, 2, .. k} */
     for ( $k = 0; $k <$V ; $k ++) 
     { 
         //Pick all vertices as source one by one 
         for ( $i = 0; $i <$V ; $i ++) 
         { 
             //Pick all vertices as destination 
             //for the above picked source 
             for ( $j = 0; $j <$V ; $j ++) 
             { 
                 //If vertex k is on the shortest path from 
                 //i to j, then update the value of dist[i][j] 
                 if ( $dist [ $i ][ $k ] + $dist [ $k ][ $j ] <
                                     $dist [ $i ][ $j ]) 
                     $dist [ $i ][ $j ] = $dist [ $i ][ $k ] +
                                     $dist [ $k ][ $j ]; 
             } 
         } 
     } 
  
     //Print the shortest distance matrix 
     printSolution( $dist , $V , $INF ); 
} 
  
/* A utility function to print solution */
function printSolution( $dist , $V , $INF ) 
{ 
     echo "The following matrix shows the " .
              "shortest distances between " . 
                 "every pair of vertices \n" ; 
     for ( $i = 0; $i <$V ; $i ++) 
     { 
         for ( $j = 0; $j <$V ; $j ++) 
         { 
             if ( $dist [ $i ][ $j ] == $INF ) 
                 echo "INF " ; 
             else
                 echo $dist [ $i ][ $j ], " " ;
         } 
         echo "\n" ; 
     } 
} 
  
//Driver Code
  
//Number of vertices in the graph 
$V = 4 ;
  
/* Define Infinite as a large enough 
value. This value will be used for
vertices not connected to each other */
$INF = 99999 ;
  
/* Let us create the following weighted graph 
         10 
(0)------->(3) 
     |     /|\ 
5 |     | 
     |     | 1 
\|/ | 
(1)------->(2) 
         3     */
$graph = array ( array (0, 5, $INF , 10), array ( $INF , 0, 3, $INF ), array ( $INF , $INF , 0, 1), array ( $INF , $INF , $INF , 0)); 
  
//Print the solution 
floydWarshall( $graph , $V , $INF ); 
  
//This code is contributed by Ryuga
?>

输出如下：

Following matrix shows the shortest distances between every pair of vertices
      0      5      8      9
    INF      0      3      4
    INF    INF      0      1
    INF    INF    INF      0

时间复杂度：O(V ^ 3)

上面的程序仅打印最短的距离。我们也可以修改解决方案以打印最短路径, 方法是将先前的信息存储在单独的2D矩阵中。

同样, 可以从limits.h中将INF的值视为INT_MAX, 以确保我们处理最大可能的值。当我们将INF作为INT_MAX时, 我们需要在上述程序中更改if条件, 以避免算术溢出。

#include 

#define INF INT_MAX
..........................
if ( dist[i][k] != INF && 
     dist[k][j] != INF && 
     dist[i][k] + dist[k][j] <dist[i][j]
    )
 dist[i][j] = dist[i][k] + dist[k][j];
...........................

如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请发表评论。

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