07 2018 档案

摘要:链接:https://www.nowcoder.com/acm/contest/141/J 题目描述 Eddy has graduated from college. Currently, he is finding his future job and a place to live. Since 阅读全文
posted @ 2018-07-31 21:24 灬从此以后灬 阅读(292) 评论(0) 推荐(0)
摘要:链接:https://www.nowcoder.com/acm/contest/141/E 题目描述 Eddy likes to play with string which is a sequence of characters. One day, Eddy has played with a s 阅读全文
posted @ 2018-07-31 17:36 灬从此以后灬 阅读(280) 评论(0) 推荐(0)
摘要:Oulipo Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 48387 Accepted: 19261 Description The French author Georges Perec (1936–1982) once w 阅读全文
posted @ 2018-07-31 15:23 灬从此以后灬 阅读(268) 评论(0) 推荐(0)
摘要:KMP算法: 一:next数组:next[i]就是前面长度为i的字符串前缀和后缀相等的最大长度,也即索引为i的字符失配时的前缀函数。 二:KMP模板 1 /* 2 pku3461(Oulipo), hdu1711(Number Sequence) 3 这个模板 字符串是从0开始的 4 Next数组是 阅读全文
posted @ 2018-07-31 13:39 灬从此以后灬 阅读(238) 评论(0) 推荐(0)
摘要:Problem A. Ascending Rating Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1310 Accepted Subm 阅读全文
posted @ 2018-07-30 21:36 灬从此以后灬 阅读(204) 评论(0) 推荐(0)
摘要:Children's Dining Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4730 Accepted: 754 Special Judge Description Usually children in kinderga 阅读全文
posted @ 2018-07-29 14:42 灬从此以后灬 阅读(234) 评论(0) 推荐(0)
摘要:The Best Path Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2104 Accepted Submission(s): 841 Pr 阅读全文
posted @ 2018-07-28 19:38 灬从此以后灬 阅读(239) 评论(0) 推荐(0)
摘要:Cover Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1543 Accepted Submission(s): 321Special Jud 阅读全文
posted @ 2018-07-27 20:53 灬从此以后灬 阅读(539) 评论(2) 推荐(0)
摘要:题目链接 https://nanti.jisuanke.com/t/28872 解析 递推 直接套杜教板子 AC代码 ........................................................................................... 阅读全文
posted @ 2018-07-25 21:34 灬从此以后灬 阅读(272) 评论(1) 推荐(0)
摘要:Ombrophobic Bovines Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20904 Accepted: 4494 Description FJ's cows really hate getting wet so m 阅读全文
posted @ 2018-07-25 21:19 灬从此以后灬 阅读(158) 评论(0) 推荐(0)
摘要:贴的某个大佬的板子,还没用过,不知道好不好用。 阅读全文
posted @ 2018-07-24 14:59 灬从此以后灬 阅读(152) 评论(0) 推荐(0)
摘要:Distinct Values Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2321 Accepted Submission(s): 748 阅读全文
posted @ 2018-07-24 14:56 灬从此以后灬 阅读(169) 评论(0) 推荐(0)
摘要:Going Home Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6478 Accepted Submission(s): 3411 Pro 阅读全文
posted @ 2018-07-23 11:35 灬从此以后灬 阅读(180) 评论(0) 推荐(0)
摘要:Farm Tour Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 19207 Accepted: 7441 Description When FJ's friends visit him on the farm, he like 阅读全文
posted @ 2018-07-22 16:17 灬从此以后灬 阅读(569) 评论(0) 推荐(0)
摘要:Just h-index Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 438 Accepted Submission(s): 203 Pr 阅读全文
posted @ 2018-07-20 21:30 灬从此以后灬 阅读(154) 评论(0) 推荐(0)
摘要:EK算法 时间复杂度o(n*m*m) 因为有反向边每次bfs时间为 n*m 每次删一条边 最多m次 代码 Dinic 时间复杂度o(n*n*m)最多计算n-1次阻塞流 每次n*m 很松的上界 代码 ISAP gap优化版 性能比dinic 好一点 ISAP算法 详解及其他版本参见https://bl 阅读全文
posted @ 2018-07-20 21:11 灬从此以后灬 阅读(282) 评论(0) 推荐(0)
摘要:链接:https://www.nowcoder.com/acm/contest/139/D来源:牛客网 同构图:假设G=(V,E)和G1=(V1,E1)是两个图,如果存在一个双射m:V→V1,使得对所有的x,y∈V均有xy∈E等价于m(x)m(y)∈E1,则称G和G1是同构的,这样的一个映射m称之为 阅读全文
posted @ 2018-07-20 16:19 灬从此以后灬 阅读(463) 评论(0) 推荐(0)
摘要:Brainman Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 12175 Accepted: 6147 Description Background Raymond Babbitt drives his brother Cha 阅读全文
posted @ 2018-07-06 13:08 灬从此以后灬 阅读(196) 评论(0) 推荐(0)
摘要:B. Nastya Studies Informatics time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output B. Nastya S 阅读全文
posted @ 2018-07-05 19:04 灬从此以后灬 阅读(239) 评论(0) 推荐(0)