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普通 ff 最大流算法可以过,但要用g++提交,我用c++ tle 了 ,思路如下:

Posted by wangjunyong at 2009-08-27 21:54:19 on Problem 2455 and last updated at 2009-08-27 21:56:06
对边排序后二分
对长度小于mid的边,在网络流中流量为 1,求最大流flow,只要flow不比要求的k条路k小,
就表示当前mid可行,其中心思想就是因为流量为1,如果要选此边,则等价于此边流量为 1。
类似求点割转求边割的思想

题目的trick是:(看以下数据)

2 2 2
1 2 2
2 1 2

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