anyone interested to discuss about the last question in ACM this year? <Tunnels
登录 | 论坛导航 -> 华新鲜事 -> 求学狮城 | 本帖共有 20 楼,当前显示第 12 楼 : 从楼主开始阅读 : 本帖树形列表 : 返回上一页
作者:房间 (等级:2 - 初出茅庐,发帖:21) 发表:2007-09-21 11:38:19  12楼 
是可以的In your case: MC(1) = 3; MC(2) = 2; MC(3) = 1; MC(4) = 2; MC(5) = 5; min cut means the minimal number of cuts to disconnect the source node from the sink node, so for node 1, we only need to cut the following edges: (1, 5), (2, 0), (4, 0). Of course there are some other minimal cuts. Accoring to my algorithm, it will output 3. According to what you have said, you want to first cut (1, 5), then the spy has only two paths to escape, namely, through node 2 and through node 4. For the path that bypasses node 2, you have to wait until the spy has entered room 2, then you need to cut (1, 2) and (2, 0), which incurs two cuts, so in total, it's three cuts. Similarly for the path through node 4. If you cut (1, 5) in the first place, and then wait until the spy has entered room 2, and you only cut (2, 0), then the spy can turn back to choose the path that goes through node 4. So you still need to cut (1, 4) or (4, 0). In this case, you again need three cuts, which matches the output of my algori
看错了
Sorry I overlooked the edge (3, 0) in your specification. So the min cuts of each node are indeed:

MC(1) = 4;
MC(2) = 2;
MC(3) = 2;
MC(4) = 2;
MC(5) = 5;
欢迎来到华新中文网,踊跃发帖是支持我们的最好方法!原文 / 传统版 / WAP版只看此人从这里展开收起列表

本帖共有 20 楼,当前显示第 12 楼,本文还有 N-1 层楼,要不你试试看:点击此处阅读更多 >>



请登录后回复:帐号   密码