Floyd Warshall算法原理和实现|DP-16

本文概述

C ++
C
Java
python
C#
的PHP

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 0589 INF034 INFINF01 INFINFINF0

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]
【Floyd Warshall算法原理和实现|DP-16】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。

文章图片
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 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)

//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 0589 INF034 INFINF01 INFINFINF0

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]; ...........................

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