https://blog.youkuaiyun.com/Revincentzh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/Revincent/article/details/84851708https://blog.youkuaiyun.com/Revincent/article/details/84851708RevincentThu, 06 Dec 2018 13:42:45 +0800https://blog.youkuaiyun.com/Revincent/article/details/72842238https://blog.youkuaiyun.com/Revincent/article/details/72842238RevincentFri, 02 Jun 2017 10:47:14 +0800https://blog.youkuaiyun.com/Revincent/article/details/70348732https://blog.youkuaiyun.com/Revincent/article/details/70348732RevincentSat, 22 Apr 2017 10:06:23 +0800https://blog.youkuaiyun.com/Revincent/article/details/70347267https://blog.youkuaiyun.com/Revincent/article/details/70347267RevincentSat, 22 Apr 2017 09:59:34 +0800https://blog.youkuaiyun.com/Revincent/article/details/69400431https://blog.youkuaiyun.com/Revincent/article/details/69400431RevincentThu, 06 Apr 2017 18:31:17 +0800https://blog.youkuaiyun.com/Revincent/article/details/69400306https://blog.youkuaiyun.com/Revincent/article/details/69400306RevincentThu, 06 Apr 2017 18:24:54 +0800b路径上的最大边权。 这是一道很裸的关于边操作的树链剖分题目，不多说了，直接上代码。 #include #include #include #define N 10005 #define inf 0x3f]]>https://blog.youkuaiyun.com/Revincent/article/details/69400218https://blog.youkuaiyun.com/Revincent/article/details/69400218RevincentThu, 06 Apr 2017 18:18:38 +0800C2之间的所有结点权值加K； 操作D：结点C1->C2之间的所有结点权值减K； 操作Q：询问结点C的值。 思路：这一眼看上去就知道是一道裸的树链剖分(点操作)，就不多说了直接上代码。没学过树链剖分的建议先入门再来看代码。 #include #inc]]>