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
作为第一步，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。想法是一一选取所有顶点并更新所有最短路径，其中包括选取的顶点作为最短路径中的中间顶点。当我们选择顶点编号 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。对于每对 (i, j) 源顶点和目标顶点，分别有两种可能的情况。
1) k 不是从 i 到 j 的最短路径中的中间顶点。我们保持 dist[i][j] 的值不变。
2) k 是从 i 到 j 的最短路径中的中间顶点。我们将 dist[i][j] 的值更新为 dist[i][k] + dist[k][j] 如果 dist[i][j] > dist[i][k] + dist[k][ j]
下图显示了上述所有对最短路径问题中的最优子结构性质。
以下是 Floyd Warshall 算法的实现。
C++
// C++ Program for Floyd Warshall Algorithm
#include
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][j] > (dist[i][k] + dist[k][j])
&& (dist[k][j] != INF
&& dist[i][k] != INF))
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 Mythri J L C
// C Program for Floyd Warshall Algorithm
#include
// 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(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 = list(map(lambda i: list(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 Mythri J LC#
// 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 MohammedPHP
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
?>Javascript
输出：
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)
上面的程序只打印最短的距离。我们还可以通过将前驱信息存储在单独的 2D 矩阵中来修改解决方案以打印最短路径。
此外，INF 的值可以作为limits.h 中的INT_MAX，以确保我们处理最大可能值。当我们取INF为INT_MAX时，我们需要改变上面程序中的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];
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