Bellman-Ford算法

解决单个最短路径问题，其中边权重可能为负，但不存在负循环。

当有向图G的某些边缘可能具有负权重时，此算法正确运行。当没有负重量的循环时，我们可以找出源与目标之间的最短路径。

它比Dijkstra的算法慢，但功能更多，因为它能够处理一些负重量边缘。

该算法检测图中的负循环并报告其存在。

基于“松弛原理”，其中更精确的值逐渐将近似值恢复到适当的距离，直到最终达到最佳解。

给定具有源s和权重函数w：E→R的加权有向图G =(V，E)，Bellman-Ford算法返回一个布尔值，该布尔值指示是否有可从源获得的负权重周期。如果存在这样的循环，该算法将产生最短的路径及其权重。当且仅当图形不包含负数(可从源到达的权重循环)时，算法才返回TRUE。

递归关系

dist ^k [u] = [min [dist ^k-1 [u]，min [dist ^k-1 [i] + cost [i，u]]]]，除u外，i等于i。

k→k是源顶点
u→u是目标顶点
i→关于顶点的待扫描边缘数。< p="">

示例：首先，我们列出所有边缘及其权重。

解：

dist ^k [u] = [min [dist ^k-1 [u]，min [dist ^k-1 [i] + cost [i，u]]]]，因为i≠u。

dist ² [2] = min [dist ¹ [2]，min [dist ¹ [1] + cost [1,2]，dist ¹ [3] + cost [3,2]，dist ¹ [4] + cost [ 4,2]，dist ¹ [5] + cost [5,2]]]

最小值= [6，0 + 6，5 +(-2)，∞+∞，∞+∞] = 3

dist ² [3] = min [dist ¹ [3]，min [dist ¹ [1] + cost [1,3]，dist ¹ [2] + cost [2,3]，dist ¹ [4] + cost [ 4,3]，dist ¹ [5] + cost [5,3]]]

最小值= [5，0 +∞，6 +∞，∞+∞，∞+∞] = 5

dist ² [4] = min [dist ¹ [4]，min [dist ¹ [1] + cost [1,4]，dist ¹ [2] + cost [2,4]，dist ¹ [3] + cost [ 3,4]，dist ¹ [5] + cost [5,4]]]

dist ² [5] = min [dist ¹ [5]，min [dist ¹ [1] + cost [1,5]，dist ¹ [2] + cost [2,5]，dist ¹ [3] + cost [ 3,5]，dist ¹ [4] + cost [4,5]]]

最小值= [∞，0 +∞，6 +∞，5 + 3，∞+ 3] = 8

dist ³ [2] = min [dist ² [2]，min [dist ² [1] + cost [1,2]，dist ² [3] + cost [3,2]，dist ² [4] + cost [ 4,2]，dist ² [5] + cost [5,2]]]

最小值= [3，0 + 6，5 +(-2)，5 +∞，8 +∞] = 3

dist ³ [3] = min [dist ² [3]，min [dist ² [1] + cost [1,3]，dist ² [2] + cost [2,3]，dist ² [4] + cost [ 4,3]，dist ² [5] + cost [5,3]]]

最小值= [5，0 +∞，3 +∞，5 +∞，8 +∞] = 5

dist ³ [4] = min [dist ² [4]，min [dist ² [1] + cost [1,4]，dist ² [2] + cost [2,4]，dist ² [3] + cost [ 3,4]，dist ² [5] + cost [5,4]]]

最小值= [5，0 +∞，3 +(-1)，5 + 4，8 +∞] = 2

dist ³ [5] = min [dist ² [5]，min [dist ² [1] + cost [1,5]，dist ² [2] + cost [2,5]，dist ² [3] + cost [ 3,5]，dist ² [4] + cost [4,5]]]

最小值= [8，0 +∞，3 +∞，5 + 3，5 + 3] = 8

dist ⁴ [2] = min [dist ³ [2]，min [dist ³ [1] + cost [1,2]，dist ³ [3] + cost [3,2]，dist ³ [4] + cost [ 4,2]，dist ³ [5] + cost [5,2]]]

最小值= [3，0 + 6，5 +(-2)，2 +∞，8 +∞] = 3

dist ⁴ [3] = min [dist ³ [3]，min [dist ³ [1] + cost [1,3]，dist ³ [2] + cost [2,3]，dist ³ [4] + cost [ 4,3]，dist ³ [5] + cost [5,3]]]

最小值= 5，0 +∞，3 +∞，2 +∞，8 +∞] = 5

dist ⁴ [4] = min [dist ³ [4]，min [dist ³ [1] + cost [1,4]，dist ³ [2] + cost [2,4]，dist ³ [3] + cost [ 3,4]，dist ³ [5] + cost [5,4]]]

最小值= [2，0 +∞，3 +(-1)，5 + 4，4，8 +∞] = 2

dist ⁴ [5] = min [dist ³ [5]，min [dist ³ [1] + cost [1,5]，dist ³ [2] + cost [2,5]，dist ³ [3] + cost [ 3,5]，dist ³ [5] + cost [4,5]]]

最小值= [8，0 +∞，3 +∞，8，5] = 5