给定一个图和图中的源顶点src ，找到从src到给定图中所有顶点的最短路径。该图可能包含负权重边缘。
我们已经讨论了Dijkstra的算法来解决这个问题。 Dijkstra的算法是Greedy算法，时间复杂度是O(VLogV)(使用Fibonacci堆)。 Dijkstra不适用于负负边的图，Bellman-Ford可用于此类图。 Bellman-Ford也比Dijkstra简单，并且非常适合分布式系统。但是Bellman-Ford的时间复杂度是O(VE)，比Dijkstra还大。
算法
以下是详细步骤。
输入：图形和源顶点src
输出： src到所有顶点的最短距离。如果存在负重量循环，则不会计算最短距离，而报告负重量循环。
1)此步骤将从源到所有顶点的距离初始化为无穷大并将与源本身的距离初始化为0。创建一个大小为| V |的数组dist []。除了dist [src](其中src是源顶点)之外的所有值都为无穷大。
2)此步骤计算最短距离。跟随| V | -1次，其中| V |是给定图中顶点的数量。
….. a)对每个边缘uv进行跟随
………………如果dist [v]> dist [u] +边缘uv的权重，则更新dist [v]
………………….dist [v] = dist [u] +边缘uv的权重
3)此步骤报告图表中是否存在负重量循环。对每个边缘uv进行跟踪
……如果dist [v]> dist [u] +边缘uv的权重，则“图形包含负权重周期”
步骤3的想法是，如果图形不包含负权重循环，则步骤2保证最短距离。如果我们再遍历所有边缘一次，并且获得任意顶点的较短路径，则负周期会变负
这是如何运作的?像其他动态编程问题一样，该算法以自底向上的方式计算最短路径。它首先计算路径中最多具有一个边缘的最短距离。然后，它计算最多具有2条边的最短路径，依此类推。在外循环的第i次迭代之后，计算出最多具有i个边的最短路径。可以有最大| V |。 –在任何简单路径中都有1条边，这就是为什么外循环运行| v |的原因– 1次。这个想法是，假设不存在负权重周期，如果我们计算出最多具有i个边缘的最短路径，那么对所有边缘进行的迭代保证给出最多具有(i + 1)个边缘的最短路径(证明很简单，您可以参考本课程或MIT视频讲座)
例子
让我们通过下面的示例图来了解算法。图像是从此来源获取的。
假设给定的源顶点为0。将所有距离初始化为无穷远，除了到源本身的距离。图中的顶点总数为5，因此所有边必须处理4次。
让所有边缘按以下顺序处理：(B，E)，(D，B)，(B，D)，(A，B)，(A，C)，(D，C)，(B，C )，(E，D)。第一次处理所有边缘时，我们得到以下距离。第一行显示初始距离。第二行显示处理边缘(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长为1个边长的所有最短路径。第二次处理所有边缘时，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长为2个边长的所有最短路径。该算法将所有边缘再处理2次。在第二次迭代之后将距离最小化，因此第三次和第四次迭代不会更新该距离。
执行：
C++
// A C++ program for Bellman-Ford's single source
// shortest path algorithm.
#include
// a structure to represent a weighted edge in graph
struct Edge {
int src, dest, weight;
};
// a structure to represent a connected, directed and
// weighted graph
struct Graph {
// V-> Number of vertices, E-> Number of edges
int V, E;
// graph is represented as an array of edges.
struct Edge* edge;
};
// Creates a graph with V vertices and E edges
struct Graph* createGraph(int V, int E)
{
struct Graph* graph = new Graph;
graph->V = V;
graph->E = E;
graph->edge = new Edge[E];
return graph;
}
// A utility function used to print the solution
void printArr(int dist[], int n)
{
printf("Vertex Distance from Source\n");
for (int i = 0; i < n; ++i)
printf("%d \t\t %d\n", i, dist[i]);
}
// The main function that finds shortest distances from src to
// all other vertices using Bellman-Ford algorithm. The function
// also detects negative weight cycle
void BellmanFord(struct Graph* graph, int src)
{
int V = graph->V;
int E = graph->E;
int dist[V];
// Step 1: Initialize distances from src to all other vertices
// as INFINITE
for (int i = 0; i < V; i++)
dist[i] = INT_MAX;
dist[src] = 0;
// Step 2: 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 = 1; i <= V - 1; i++) {
for (int j = 0; j < E; j++) {
int u = graph->edge[j].src;
int v = graph->edge[j].dest;
int weight = graph->edge[j].weight;
if (dist[u] != INT_MAX && dist[u] + weight < dist[v])
dist[v] = dist[u] + weight;
}
}
// Step 3: 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 u = graph->edge[i].src;
int v = graph->edge[i].dest;
int weight = graph->edge[i].weight;
if (dist[u] != INT_MAX && dist[u] + weight < dist[v]) {
printf("Graph contains negative weight cycle");
return; // If negative cycle is detected, simply return
}
}
printArr(dist, V);
return;
}
// Driver program to test above functions
int main()
{
/* Let us create the graph given in above example */
int V = 5; // Number of vertices in graph
int E = 8; // Number of edges in graph
struct Graph* graph = createGraph(V, E);
// add edge 0-1 (or A-B in above figure)
graph->edge[0].src = 0;
graph->edge[0].dest = 1;
graph->edge[0].weight = -1;
// add edge 0-2 (or A-C in above figure)
graph->edge[1].src = 0;
graph->edge[1].dest = 2;
graph->edge[1].weight = 4;
// add edge 1-2 (or B-C in above figure)
graph->edge[2].src = 1;
graph->edge[2].dest = 2;
graph->edge[2].weight = 3;
// add edge 1-3 (or B-D in above figure)
graph->edge[3].src = 1;
graph->edge[3].dest = 3;
graph->edge[3].weight = 2;
// add edge 1-4 (or A-E in above figure)
graph->edge[4].src = 1;
graph->edge[4].dest = 4;
graph->edge[4].weight = 2;
// add edge 3-2 (or D-C in above figure)
graph->edge[5].src = 3;
graph->edge[5].dest = 2;
graph->edge[5].weight = 5;
// add edge 3-1 (or D-B in above figure)
graph->edge[6].src = 3;
graph->edge[6].dest = 1;
graph->edge[6].weight = 1;
// add edge 4-3 (or E-D in above figure)
graph->edge[7].src = 4;
graph->edge[7].dest = 3;
graph->edge[7].weight = -3;
BellmanFord(graph, 0);
return 0;
} Java
// A Java program for Bellman-Ford's single source shortest path
// algorithm.
import java.util.*;
import java.lang.*;
import java.io.*;
// A class to represent a connected, directed and weighted graph
class Graph {
// A class to represent a weighted edge in graph
class Edge {
int src, dest, weight;
Edge()
{
src = dest = weight = 0;
}
};
int V, E;
Edge edge[];
// Creates a graph with V vertices and E edges
Graph(int v, int e)
{
V = v;
E = e;
edge = new Edge[e];
for (int i = 0; i < e; ++i)
edge[i] = new Edge();
}
// The main function that finds shortest distances from src
// to all other vertices using Bellman-Ford algorithm. The
// function also detects negative weight cycle
void BellmanFord(Graph graph, int src)
{
int V = graph.V, E = graph.E;
int dist[] = new int[V];
// Step 1: Initialize distances from src to all other
// vertices as INFINITE
for (int i = 0; i < V; ++i)
dist[i] = Integer.MAX_VALUE;
dist[src] = 0;
// Step 2: 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 = 1; i < V; ++i) {
for (int j = 0; j < E; ++j) {
int u = graph.edge[j].src;
int v = graph.edge[j].dest;
int weight = graph.edge[j].weight;
if (dist[u] != Integer.MAX_VALUE && dist[u] + weight < dist[v])
dist[v] = dist[u] + weight;
}
}
// Step 3: 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 j = 0; j < E; ++j) {
int u = graph.edge[j].src;
int v = graph.edge[j].dest;
int weight = graph.edge[j].weight;
if (dist[u] != Integer.MAX_VALUE && dist[u] + weight < dist[v]) {
System.out.println("Graph contains negative weight cycle");
return;
}
}
printArr(dist, V);
}
// A utility function used to print the solution
void printArr(int dist[], int V)
{
System.out.println("Vertex Distance from Source");
for (int i = 0; i < V; ++i)
System.out.println(i + "\t\t" + dist[i]);
}
// Driver method to test above function
public static void main(String[] args)
{
int V = 5; // Number of vertices in graph
int E = 8; // Number of edges in graph
Graph graph = new Graph(V, E);
// add edge 0-1 (or A-B in above figure)
graph.edge[0].src = 0;
graph.edge[0].dest = 1;
graph.edge[0].weight = -1;
// add edge 0-2 (or A-C in above figure)
graph.edge[1].src = 0;
graph.edge[1].dest = 2;
graph.edge[1].weight = 4;
// add edge 1-2 (or B-C in above figure)
graph.edge[2].src = 1;
graph.edge[2].dest = 2;
graph.edge[2].weight = 3;
// add edge 1-3 (or B-D in above figure)
graph.edge[3].src = 1;
graph.edge[3].dest = 3;
graph.edge[3].weight = 2;
// add edge 1-4 (or A-E in above figure)
graph.edge[4].src = 1;
graph.edge[4].dest = 4;
graph.edge[4].weight = 2;
// add edge 3-2 (or D-C in above figure)
graph.edge[5].src = 3;
graph.edge[5].dest = 2;
graph.edge[5].weight = 5;
// add edge 3-1 (or D-B in above figure)
graph.edge[6].src = 3;
graph.edge[6].dest = 1;
graph.edge[6].weight = 1;
// add edge 4-3 (or E-D in above figure)
graph.edge[7].src = 4;
graph.edge[7].dest = 3;
graph.edge[7].weight = -3;
graph.BellmanFord(graph, 0);
}
}
// Contributed by Aakash HasijaPython3
# Python3 program for Bellman-Ford's single source
# shortest path algorithm.
# Class to represent a graph
class Graph:
def __init__(self, vertices):
self.V = vertices # No. of vertices
self.graph = []
# function to add an edge to graph
def addEdge(self, u, v, w):
self.graph.append([u, v, w])
# utility function used to print the solution
def printArr(self, dist):
print("Vertex Distance from Source")
for i in range(self.V):
print("{0}\t\t{1}".format(i, dist[i]))
# The main function that finds shortest distances from src to
# all other vertices using Bellman-Ford algorithm. The function
# also detects negative weight cycle
def BellmanFord(self, src):
# Step 1: Initialize distances from src to all other vertices
# as INFINITE
dist = [float("Inf")] * self.V
dist[src] = 0
# Step 2: Relax all edges |V| - 1 times. A simple shortest
# path from src to any other vertex can have at-most |V| - 1
# edges
for _ in range(self.V - 1):
# Update dist value and parent index of the adjacent vertices of
# the picked vertex. Consider only those vertices which are still in
# queue
for u, v, w in self.graph:
if dist[u] != float("Inf") and dist[u] + w < dist[v]:
dist[v] = dist[u] + w
# Step 3: 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 u, v, w in self.graph:
if dist[u] != float("Inf") and dist[u] + w < dist[v]:
print("Graph contains negative weight cycle")
return
# print all distance
self.printArr(dist)
g = Graph(5)
g.addEdge(0, 1, -1)
g.addEdge(0, 2, 4)
g.addEdge(1, 2, 3)
g.addEdge(1, 3, 2)
g.addEdge(1, 4, 2)
g.addEdge(3, 2, 5)
g.addEdge(3, 1, 1)
g.addEdge(4, 3, -3)
# Print the solution
g.BellmanFord(0)
# Initially, Contributed by Neelam Yadav
# Later On, Edited by Himanshu GargC#
// A C# program for Bellman-Ford's single source shortest path
// algorithm.
using System;
// A class to represent a connected, directed and weighted graph
class Graph {
// A class to represent a weighted edge in graph
class Edge {
public int src, dest, weight;
public Edge()
{
src = dest = weight = 0;
}
};
int V, E;
Edge[] edge;
// Creates a graph with V vertices and E edges
Graph(int v, int e)
{
V = v;
E = e;
edge = new Edge[e];
for (int i = 0; i < e; ++i)
edge[i] = new Edge();
}
// The main function that finds shortest distances from src
// to all other vertices using Bellman-Ford algorithm. The
// function also detects negative weight cycle
void BellmanFord(Graph graph, int src)
{
int V = graph.V, E = graph.E;
int[] dist = new int[V];
// Step 1: Initialize distances from src to all other
// vertices as INFINITE
for (int i = 0; i < V; ++i)
dist[i] = int.MaxValue;
dist[src] = 0;
// Step 2: 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 = 1; i < V; ++i) {
for (int j = 0; j < E; ++j) {
int u = graph.edge[j].src;
int v = graph.edge[j].dest;
int weight = graph.edge[j].weight;
if (dist[u] != int.MaxValue && dist[u] + weight < dist[v])
dist[v] = dist[u] + weight;
}
}
// Step 3: 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 j = 0; j < E; ++j) {
int u = graph.edge[j].src;
int v = graph.edge[j].dest;
int weight = graph.edge[j].weight;
if (dist[u] != int.MaxValue && dist[u] + weight < dist[v]) {
Console.WriteLine("Graph contains negative weight cycle");
return;
}
}
printArr(dist, V);
}
// A utility function used to print the solution
void printArr(int[] dist, int V)
{
Console.WriteLine("Vertex Distance from Source");
for (int i = 0; i < V; ++i)
Console.WriteLine(i + "\t\t" + dist[i]);
}
// Driver method to test above function
public static void Main()
{
int V = 5; // Number of vertices in graph
int E = 8; // Number of edges in graph
Graph graph = new Graph(V, E);
// add edge 0-1 (or A-B in above figure)
graph.edge[0].src = 0;
graph.edge[0].dest = 1;
graph.edge[0].weight = -1;
// add edge 0-2 (or A-C in above figure)
graph.edge[1].src = 0;
graph.edge[1].dest = 2;
graph.edge[1].weight = 4;
// add edge 1-2 (or B-C in above figure)
graph.edge[2].src = 1;
graph.edge[2].dest = 2;
graph.edge[2].weight = 3;
// add edge 1-3 (or B-D in above figure)
graph.edge[3].src = 1;
graph.edge[3].dest = 3;
graph.edge[3].weight = 2;
// add edge 1-4 (or A-E in above figure)
graph.edge[4].src = 1;
graph.edge[4].dest = 4;
graph.edge[4].weight = 2;
// add edge 3-2 (or D-C in above figure)
graph.edge[5].src = 3;
graph.edge[5].dest = 2;
graph.edge[5].weight = 5;
// add edge 3-1 (or D-B in above figure)
graph.edge[6].src = 3;
graph.edge[6].dest = 1;
graph.edge[6].weight = 1;
// add edge 4-3 (or E-D in above figure)
graph.edge[7].src = 4;
graph.edge[7].dest = 3;
graph.edge[7].weight = -3;
graph.BellmanFord(graph, 0);
}
// This code is contributed by Ryuga
}输出：
Vertex Distance from Source
0 0
1 -1
2 2
3 -2
4 1笔记
1)在图表的各种应用中都可以找到负权重。例如，如果不遵循路径的成本，那么遵循该路径可能会获得一些好处。
2) Bellman-Ford在分布式系统上工作得更好(比Dijksra更好)。与Dijkstra的需要找到所有顶点的最小值的方法不同，在Bellman-Ford中，边被一一视为。
锻炼
1)标准的Bellman-Ford算法仅在没有负权重循环的情况下报告最短路径。对其进行修改，以便即使负重周期也能报告最小距离。
2)我们可以使用Dijkstra的算法为负权重的图提供最短路径吗?一个想法是，计算最小权重值，对所有权重添加一个正值(等于最小权重值的绝对值)，然后运行Dijkstra的算法来计算权重。修改后的图。这个算法会起作用吗?
Bellman Ford算法(简单实现)
参考：
http://www.youtube.com/watch?v=Ttezuzs39nk
http://en.wikipedia.org/wiki/Bellman%E2%80%93Ford_algorithm
http://www.cs.arizona.edu/classes/cs445/spring07/ShortestPath2.prn.pdf