https://blog.youkuaiyun.com/Ano_pathzh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/Ano_path/article/details/52201482https://blog.youkuaiyun.com/Ano_path/article/details/52201482Ano_pathSat, 13 Aug 2016 20:43:55 +0800https://blog.youkuaiyun.com/Ano_path/article/details/52143367https://blog.youkuaiyun.com/Ano_path/article/details/52143367Ano_pathSun, 07 Aug 2016 16:48:44 +0800https://blog.youkuaiyun.com/Ano_path/article/details/52049025https://blog.youkuaiyun.com/Ano_path/article/details/52049025Ano_pathWed, 27 Jul 2016 20:10:45 +0800https://blog.youkuaiyun.com/Ano_path/article/details/52026945https://blog.youkuaiyun.com/Ano_path/article/details/52026945Ano_pathMon, 25 Jul 2016 19:54:15 +0800https://blog.youkuaiyun.com/Ano_path/article/details/52016016https://blog.youkuaiyun.com/Ano_path/article/details/52016016Ano_pathSun, 24 Jul 2016 20:31:01 +0800https://blog.youkuaiyun.com/Ano_path/article/details/51974113https://blog.youkuaiyun.com/Ano_path/article/details/51974113Ano_pathWed, 20 Jul 2016 23:02:38 +0800https://blog.youkuaiyun.com/Ano_path/article/details/51945128https://blog.youkuaiyun.com/Ano_path/article/details/51945128Ano_pathMon, 18 Jul 2016 20:18:15 +0800https://blog.youkuaiyun.com/Ano_path/article/details/51945011https://blog.youkuaiyun.com/Ano_path/article/details/51945011Ano_pathMon, 18 Jul 2016 20:05:20 +0800https://blog.youkuaiyun.com/Ano_path/article/details/51927068https://blog.youkuaiyun.com/Ano_path/article/details/51927068Ano_pathSat, 16 Jul 2016 23:22:07 +0800https://blog.youkuaiyun.com/Ano_path/article/details/51926561https://blog.youkuaiyun.com/Ano_path/article/details/51926561Ano_pathSat, 16 Jul 2016 21:35:22 +0800=n/2，那么就一定存在一个方案，否则就无解。 显然，当存在一个帮派的数量大于n/2时，剩下的奶牛都不可能将它们全部抵消。 基于这个思想，就不难得出答案： 我们设所剩奶牛数为ans，那么首先ans要满足除了1号以外的每个帮派内的奶牛数的2倍都小于n-ans+1（即预留了an]]>https://blog.youkuaiyun.com/Ano_path/article/details/51919923https://blog.youkuaiyun.com/Ano_path/article/details/51919923Ano_pathFri, 15 Jul 2016 17:51:15 +0800=k 2.R-L最小满足1的不一定满足2，但是满足2的一定要满足1 所以我们考虑枚举L和R，再检验是否满足sum>=k最暴力的做法是： 将边按照频率的大小排序，枚举一维下界，枚举一维上界，将频率在区间内的边的两端点合并，通过并查集维护每个连通块的大小，效率是O(m3m^3)，不能通过所有数据点。考虑枚举上界的情况： 由于下界不变，枚举上]]>https://blog.youkuaiyun.com/Ano_path/article/details/51913299https://blog.youkuaiyun.com/Ano_path/article/details/51913299Ano_pathThu, 14 Jul 2016 21:59:10 +0800