https://blog.csdn.net/python_yydzh-cnhttps://blog.csdn.net/https://blog.csdn.net/python_yyd/article/details/152723958https://blog.csdn.net/python_yyd/article/details/152723958python_yydWed, 08 Oct 2025 15:19:59 +0800= 2 * (n - 1) ，解出这个式子得到 n <= 4，这就代表我们只需要去找小于等于4的连通块就可以了。]]>https://blog.csdn.net/python_yyd/article/details/152607510https://blog.csdn.net/python_yyd/article/details/152607510python_yydMon, 06 Oct 2025 18:21:29 +0800https://blog.csdn.net/python_yyd/article/details/152422886https://blog.csdn.net/python_yyd/article/details/152422886python_yydSat, 04 Oct 2025 13:03:31 +0800https://blog.csdn.net/python_yyd/article/details/152414322https://blog.csdn.net/python_yyd/article/details/152414322python_yydThu, 02 Oct 2025 17:52:11 +0800https://blog.csdn.net/python_yyd/article/details/152272918https://blog.csdn.net/python_yyd/article/details/152272918python_yydTue, 30 Sep 2025 00:29:54 +0800=sum，算出来的结果是c （这个c一定是某几个最大的正数的和， 剩下的整数就是会被抵消掉的正数和，|负数和| >= 没有选的正数）， 正数的个数 - c + 1就是可以被抵消的正数和，也就是最后的结果。分析：读题感觉和叶子节点有关，也就是非叶子节点的个数，非叶子节点的个数就等于操作的总次数，所以判断奇偶直接输出即可。分析：bfs求最短路并且要记录状态，不能贪心，因为没有办法模拟可能右边往左边运羊的情况。]]>https://blog.csdn.net/python_yyd/article/details/152192218https://blog.csdn.net/python_yyd/article/details/152192218python_yydSun, 28 Sep 2025 12:08:53 +0800https://blog.csdn.net/python_yyd/article/details/152187746https://blog.csdn.net/python_yyd/article/details/152187746python_yydSun, 28 Sep 2025 00:53:09 +0800= (n + 1) / 2;分析：手速可以铜牌的题，本身不难，但是要注意很多细节，以及必须按照dfs的顺序进行，不能手动模拟。]]>https://blog.csdn.net/python_yyd/article/details/152129977https://blog.csdn.net/python_yyd/article/details/152129977python_yydFri, 26 Sep 2025 20:21:06 +0800https://blog.csdn.net/python_yyd/article/details/152093833https://blog.csdn.net/python_yyd/article/details/152093833python_yydFri, 26 Sep 2025 12:30:35 +0800https://blog.csdn.net/python_yyd/article/details/152045920https://blog.csdn.net/python_yyd/article/details/152045920python_yydWed, 24 Sep 2025 17:41:03 +0800https://blog.csdn.net/python_yyd/article/details/151859045https://blog.csdn.net/python_yyd/article/details/151859045python_yydFri, 19 Sep 2025 11:52:17 +080026退出即可，最后搜到的最小值就是我们要的答案。]]>https://blog.csdn.net/python_yyd/article/details/150398336https://blog.csdn.net/python_yyd/article/details/150398336python_yydThu, 14 Aug 2025 20:11:24 +0800https://blog.csdn.net/python_yyd/article/details/150354714https://blog.csdn.net/python_yyd/article/details/150354714python_yydWed, 13 Aug 2025 21:45:19 +0800https://blog.csdn.net/python_yyd/article/details/150346697https://blog.csdn.net/python_yyd/article/details/150346697python_yydWed, 13 Aug 2025 15:41:49 +0800https://blog.csdn.net/python_yyd/article/details/150343826https://blog.csdn.net/python_yyd/article/details/150343826python_yydWed, 13 Aug 2025 14:36:14 +0800https://blog.csdn.net/python_yyd/article/details/148631553https://blog.csdn.net/python_yyd/article/details/148631553python_yydMon, 28 Jul 2025 11:36:26 +0800https://blog.csdn.net/python_yyd/article/details/147000494https://blog.csdn.net/python_yyd/article/details/147000494python_yydFri, 04 Apr 2025 20:00:34 +0800https://blog.csdn.net/python_yyd/article/details/146944631https://blog.csdn.net/python_yyd/article/details/146944631python_yydThu, 03 Apr 2025 01:51:43 +0800https://blog.csdn.net/python_yyd/article/details/146636184https://blog.csdn.net/python_yyd/article/details/146636184python_yydSat, 29 Mar 2025 01:03:21 +0800https://blog.csdn.net/python_yyd/article/details/146620891https://blog.csdn.net/python_yyd/article/details/146620891python_yydFri, 28 Mar 2025 22:57:29 +0800