Order to map any of the distance between two nodes, (2) with two-dimensional array of storage C, A, C is the original cost matrix, A is to find the distance matrix (3) The algorithm uses the triple loop, one of the most outer loop variable must be representative of intermediate nodes, the middle loop variable represents the first node and the inner loop variable represents the end nodes. (4) tried to three-loop order of the variables to make some changes, the most outer loop variables still represent intermediate nodes, while the middle loop variable represents the end nodes, the inner loop variable represents the first node. The two approaches to compare the results to see whether the results the same (5) shows the results must be clear and understandable (6) The question running results

shuanfa2\test\A.txt
........\....\APSP.out
........\....\APSP.txt
........\....\cost.txt
........\....\Debug\test.exe
........\....\.....\test.ilk
........\....\.....\test.obj
........\....\.....\test.pch
........\....\.....\test.pdb
........\....\.....\vc60.idb
........\....\.....\vc60.pdb
........\....\test.cpp
........\....\test.dsp
........\....\test.dsw
........\....\test.ncb
........\....\test.opt
........\....\test.plg
........\....\Debug
........\test
shuanfa2

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