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修正卡码网0095.城市间货物运输II中,Python解决SPFA方法求解含有负回路的最短路问题的错误

原代码(错误的)

                count[next_node] += 1
                if next_node not in d:
                      d.append(next_node)

在输入例子

4 6
1 4 3
1 2 1
1 3 1
3 2 -2
2 4 1
3 4 0

会输出circle,但正确结果是0

纠正代码(可以通过全部例子)

                if next_node not in d:
                      count[next_node] += 1
                      d.append(next_node)

修正卡码网0095.城市间货物运输II中,Python解决SPFA方法求解含有负回路的最短路问题的错误

count[next_node] += 1应该在 if next_node not in d: 语句块内
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