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Re:这样做怎么就不行呢?:先求出所有强连通分支,再对每个点判断它所指向的每一点和它自己是否在同一SCC中,是则为sink点In Reply To:这样做怎么就不行呢?:先求出所有强连通分支,再对每个点判断它所指向的每一点和它自己是否在同一SCC中,是则为sink点 Posted by:006794 at 2007-02-24 18:35:13 如果有一条边从A连通分量指向B连通分量，那么A里所有的点都不行。 Followed by: Post your reply here: |
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