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Re:将i拆为i->i',容量为m[i].再加一个源点s,s->i(0~n-1)为p[i].然后对每个点求最大流.In Reply To:Re:用网络流做该如何建图？ Posted by:lxc0601 at 2008-08-24 13:54:21 > 拆点？ 不过我的tle 。。 将i拆为i->i',容量为m[i].再加一个源点s,s->i(0~n-1)容量为p[i].然后对每个点求最大流. 用Dinic很快的. Followed by: Post your reply here: |
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