我们已经介绍了 Bellman Ford 并在这里讨论了实施。
输入:图形和源顶点src
输出:从src到所有顶点的最短距离。如果存在负重量周期,则不计算最短距离,报告负重量周期。
1)这一步将源到所有顶点的距离初始化为无穷大,到源自身的距离初始化为 0。创建一个大小为 |V| 的数组 dist[]除了 dist[src] 其中 src 是源顶点之外,所有值都为无穷大。
2)这一步计算最短距离。跟随 |V|-1 次,其中 |V|是给定图中的顶点数。
….. a)对每个边 uv 执行以下操作
………………如果dist[v] > dist[u] + 边uv的权重,则更新dist[v]
………………….dist[v] = dist[u] + 边 uv 的权重
3)此步骤报告图中是否存在负权重循环。对每个边缘 uv 执行以下操作
……如果 dist[v] > dist[u] + 边 uv 的权重,则“图包含负权重循环”
第 3 步的想法是,如果图不包含负权重循环,则第 2 步保证最短距离。如果我们再一次遍历所有边并为任何顶点获得一条更短的路径,那么存在负权重循环
例子
让我们通过以下示例图来理解算法。图像取自此来源。
让给定的源顶点为 0。将所有距离初始化为无穷大,除了到源本身的距离。图中的顶点总数为 5,因此所有边都必须处理 4 次。
让所有边按以下顺序处理:(B, E), (D, B), (B, D), (A, B), (A, C), (D, C), (B, C) , (E, D)。第一次处理所有边时,我们得到以下距离。中的第一行显示初始距离。第二行显示处理边 (B, E)、(D, B)、(B, D) 和 (A, B) 时的距离。第三行显示处理 (A, C) 时的距离。第四行显示何时处理(D,C),(B,C)和(E,D)。
第一次迭代保证给出最多 1 条边长的所有最短路径。当第二次处理所有边时,我们得到以下距离(最后一行显示最终值)。
第二次迭代保证给出最多 2 条边长的所有最短路径。该算法将所有边再处理 2 次。第二次迭代后距离最小化,因此第三次和第四次迭代不会更新距离。
C++
// A C++ program for Bellman-Ford's single source
// shortest path algorithm.
#include
using namespace std;
// The main function that finds shortest
// distances from src to all other vertices
// using Bellman-Ford algorithm. The function
// also detects negative weight cycle
// The row graph[i] represents i-th edge with
// three values u, v and w.
void BellmanFord(int graph[][3], int V, int E,
int src)
{
// Initialize distance of all vertices as infinite.
int dis[V];
for (int i = 0; i < V; i++)
dis[i] = INT_MAX;
// initialize distance of source as 0
dis[src] = 0;
// Relax all edges |V| - 1 times. A simple
// shortest path from src to any other
// vertex can have at-most |V| - 1 edges
for (int i = 0; i < V - 1; i++) {
for (int j = 0; j < E; j++) {
if (dis[graph[j][0]] != INT_MAX && dis[graph[j][0]] + graph[j][2] <
dis[graph[j][1]])
dis[graph[j][1]] =
dis[graph[j][0]] + graph[j][2];
}
}
// check for negative-weight cycles.
// The above step guarantees shortest
// distances if graph doesn't contain
// negative weight cycle. If we get a
// shorter path, then there is a cycle.
for (int i = 0; i < E; i++) {
int x = graph[i][0];
int y = graph[i][1];
int weight = graph[i][2];
if (dis[x] != INT_MAX &&
dis[x] + weight < dis[y])
cout << "Graph contains negative"
" weight cycle"
<< endl;
}
cout << "Vertex Distance from Source" << endl;
for (int i = 0; i < V; i++)
cout << i << "\t\t" << dis[i] << endl;
}
// Driver program to test above functions
int main()
{
int V = 5; // Number of vertices in graph
int E = 8; // Number of edges in graph
// Every edge has three values (u, v, w) where
// the edge is from vertex u to v. And weight
// of the edge is w.
int graph[][3] = { { 0, 1, -1 }, { 0, 2, 4 },
{ 1, 2, 3 }, { 1, 3, 2 },
{ 1, 4, 2 }, { 3, 2, 5 },
{ 3, 1, 1 }, { 4, 3, -3 } };
BellmanFord(graph, V, E, 0);
return 0;
}
Java
// A Java program for Bellman-Ford's single source
// shortest path algorithm.
class GFG
{
// The main function that finds shortest
// distances from src to all other vertices
// using Bellman-Ford algorithm. The function
// also detects negative weight cycle
// The row graph[i] represents i-th edge with
// three values u, v and w.
static void BellmanFord(int graph[][], int V, int E,
int src)
{
// Initialize distance of all vertices as infinite.
int []dis = new int[V];
for (int i = 0; i < V; i++)
dis[i] = Integer.MAX_VALUE;
// initialize distance of source as 0
dis[src] = 0;
// Relax all edges |V| - 1 times. A simple
// shortest path from src to any other
// vertex can have at-most |V| - 1 edges
for (int i = 0; i < V - 1; i++)
{
for (int j = 0; j < E; j++)
{
if (dis[graph[j][0]] != Integer.MAX_VALUE && dis[graph[j][0]] + graph[j][2] <
dis[graph[j][1]])
dis[graph[j][1]] =
dis[graph[j][0]] + graph[j][2];
}
}
// check for negative-weight cycles.
// The above step guarantees shortest
// distances if graph doesn't contain
// negative weight cycle. If we get a
// shorter path, then there is a cycle.
for (int i = 0; i < E; i++)
{
int x = graph[i][0];
int y = graph[i][1];
int weight = graph[i][2];
if (dis[x] != Integer.MAX_VALUE &&
dis[x] + weight < dis[y])
System.out.println("Graph contains negative"
+" weight cycle");
}
System.out.println("Vertex Distance from Source");
for (int i = 0; i < V; i++)
System.out.println(i + "\t\t" + dis[i]);
}
// Driver code
public static void main(String[] args)
{
int V = 5; // Number of vertices in graph
int E = 8; // Number of edges in graph
// Every edge has three values (u, v, w) where
// the edge is from vertex u to v. And weight
// of the edge is w.
int graph[][] = { { 0, 1, -1 }, { 0, 2, 4 },
{ 1, 2, 3 }, { 1, 3, 2 },
{ 1, 4, 2 }, { 3, 2, 5 },
{ 3, 1, 1 }, { 4, 3, -3 } };
BellmanFord(graph, V, E, 0);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program for Bellman-Ford's
# single source shortest path algorithm.
from sys import maxsize
# The main function that finds shortest
# distances from src to all other vertices
# using Bellman-Ford algorithm. The function
# also detects negative weight cycle
# The row graph[i] represents i-th edge with
# three values u, v and w.
def BellmanFord(graph, V, E, src):
# Initialize distance of all vertices as infinite.
dis = [maxsize] * V
# initialize distance of source as 0
dis[src] = 0
# Relax all edges |V| - 1 times. A simple
# shortest path from src to any other
# vertex can have at-most |V| - 1 edges
for i in range(V - 1):
for j in range(E):
if dis[graph[j][0]] + \
graph[j][2] < dis[graph[j][1]]:
dis[graph[j][1]] = dis[graph[j][0]] + \
graph[j][2]
# check for negative-weight cycles.
# The above step guarantees shortest
# distances if graph doesn't contain
# negative weight cycle. If we get a
# shorter path, then there is a cycle.
for i in range(E):
x = graph[i][0]
y = graph[i][1]
weight = graph[i][2]
if dis[x] != maxsize and dis[x] + \
weight < dis[y]:
print("Graph contains negative weight cycle")
print("Vertex Distance from Source")
for i in range(V):
print("%d\t\t%d" % (i, dis[i]))
# Driver Code
if __name__ == "__main__":
V = 5 # Number of vertices in graph
E = 8 # Number of edges in graph
# Every edge has three values (u, v, w) where
# the edge is from vertex u to v. And weight
# of the edge is w.
graph = [[0, 1, -1], [0, 2, 4], [1, 2, 3],
[1, 3, 2], [1, 4, 2], [3, 2, 5],
[3, 1, 1], [4, 3, -3]]
BellmanFord(graph, V, E, 0)
# This code is contributed by
# sanjeev2552
C#
// C# program for Bellman-Ford's single source
// shortest path algorithm.
using System;
class GFG
{
// The main function that finds shortest
// distances from src to all other vertices
// using Bellman-Ford algorithm. The function
// also detects negative weight cycle
// The row graph[i] represents i-th edge with
// three values u, v and w.
static void BellmanFord(int [,]graph, int V,
int E, int src)
{
// Initialize distance of all vertices as infinite.
int []dis = new int[V];
for (int i = 0; i < V; i++)
dis[i] = int.MaxValue;
// initialize distance of source as 0
dis[src] = 0;
// Relax all edges |V| - 1 times. A simple
// shortest path from src to any other
// vertex can have at-most |V| - 1 edges
for (int i = 0; i < V - 1; i++)
{
for (int j = 0; j < E; j++)
{
if (dis[graph[j, 0]] = int.MaxValue && dis[graph[j, 0]] + graph[j, 2] <
dis[graph[j, 1]])
dis[graph[j, 1]] =
dis[graph[j, 0]] + graph[j, 2];
}
}
// check for negative-weight cycles.
// The above step guarantees shortest
// distances if graph doesn't contain
// negative weight cycle. If we get a
// shorter path, then there is a cycle.
for (int i = 0; i < E; i++)
{
int x = graph[i, 0];
int y = graph[i, 1];
int weight = graph[i, 2];
if (dis[x] != int.MaxValue &&
dis[x] + weight < dis[y])
Console.WriteLine("Graph contains negative" +
" weight cycle");
}
Console.WriteLine("Vertex Distance from Source");
for (int i = 0; i < V; i++)
Console.WriteLine(i + "\t\t" + dis[i]);
}
// Driver code
public static void Main(String[] args)
{
int V = 5; // Number of vertices in graph
int E = 8; // Number of edges in graph
// Every edge has three values (u, v, w) where
// the edge is from vertex u to v. And weight
// of the edge is w.
int [,]graph = {{ 0, 1, -1 }, { 0, 2, 4 },
{ 1, 2, 3 }, { 1, 3, 2 },
{ 1, 4, 2 }, { 3, 2, 5 },
{ 3, 1, 1 }, { 4, 3, -3 }};
BellmanFord(graph, V, E, 0);
}
}
// This code is contributed by Princi Singh
PHP
Javascript
Vertex Distance from Source
0 0
1 -1
2 2
3 -2
4 1
时间复杂度: O(VE)
此实现由PrateekGupta10建议
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