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思路以K为根建树，遍历找出必须要经过的边，得边权和为sum， 要使dis最短，则有一点可以不返回，那点不就是要到的点 中离K最远的city吗？记为MAX，则mindis=sum*2-MAX。 一开始TLE 优化了下之后发现是WA。。。想不到招了 Followed by:
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