1 void Floyd(int *edge, int *path, const int order, const int maxNum)
2 {
3 int length;
4 if (path != NULL)
5 for (int i = 0; i < order; i++)
6 for (int j = 0; j < order; j++)
7 if (i != j && (*(edge+ order * i + j) < maxNum))
8 *(path + order * i + j) = i;
9 else
10 *(path + order * i + j) = 0;
11
12 for (int k = 0; k < order; k++)
13 for (int i = 0; i < order; i++)
14 for (int j = 0; j < order; j++)
15 {
16 length = (*(edge + i * order + k) + *(edge + k * order + j));
17 if (length < *(edge + i * order + j))
18 {
19 *(edge + i * order + j) = length;
20 if (path != NULL)
21 *(path + i * order + j) = *(path + k * order + j);
22 }
23 }
24 }
调用示例:
1 cosnt int MaxNum = 1000;
2 int a[4][4] = {{0, 1, MaxNum, 4}, {MaxNum, 0, 9, 2}, {3, 5, 0, 8}, {MaxNum, MaxNum, 6, 0}};
3 int b[4][4];
4 Floyd((int *)a, (int *)b, 4, MaxNum);
