1.求所有点对的最短路径问题，设G=(V,E)是一个有向图，其中的每条边(i,j)由一个非负的长度l[i,j]，如果从顶点i到顶点j没有边，则l[i,j]=∞。要找出从每个顶点到其他所有顶点的距离，这里从顶点x到顶点y的距离是指从x到y的最短路径的长度。

2. 通过一个图的权值矩阵求出它的每两点间的最短路径矩阵。

3. 从图的带权邻接矩阵A=[a(i,j)] n×n开始，递归地进行n次更新，即由矩阵D(0)=A，按一个公式，构造出矩阵D(1)；又用同样地公式由D(1)构造出D(2)；……；最后又用同样的公式由D(n-1)构造出矩阵D(n)。矩阵D(n)的i行j列元素便是i号顶点到j号顶点的最短路径长度，称D(n)为图的距离矩阵，同时还可引入一个后继节点矩阵path来记录两点间的最短路径。-Find all points on the shortest path problem, let G = (V, E) be a directed graph, where each edge (i, j) by a non-negative length l [i, j], if from the vertex i no edge to the vertex j, then l [i, j] = ∞. To find out from each vertex to all other nodes in the distance, where the vertices from vertex x to y is the distance from x to y represents the shortest path length.

(2) the weight by a graph in which each matrix calculated shortest path between two matrices.

3 from the graph adjacency matrix with the right A = [a (i, j)] n × n starts, recursively updated n times, i.e. by the matrix D (0) = A, according to a formula is constructed matrix D (1)　also used the same way by the formula D (1) construct a D (2)　......　finally use the same formula consists of D (n-1) construct a matrix D (n). Matrix D (n) of the i-th row j-th column is the i-th element of the j-th vertex to vertex of the shortest path length, called D (n) of the distance matrix of Fig, while also introducing a successo