求一个图 变成生成树至少还要加几条边(成环的边要删掉,但不用统计) Sample Input4 2 //n m1 3//u v4 33 31 21 32 35 21 23 5999 00 Sample Output102998 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # include &…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213 Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other,…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1232 畅通工程 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 65589    Accepted Submission(s): 35027 Problem Description 某省调查城镇交通状况,得到现有城镇道路统计表,表中列出了每…

求连通分量 Sample Input2 //T5 3 //n m1 2// u v2 34 5 5 12 5 Sample Output24 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # include <queue> # define LL long long using na…

http://acm.hdu.edu.cn/showproblem.php?pid=1213 题意: 这个问题的一个重要规则是,如果我告诉你A知道B,B知道C,这意味着A,B,C知道对方,所以他们可以留在一个桌子. 例如:如果我告诉你A知道B,B知道C,D知道E,所以A,B,C可以留在一个桌子中,D,E必须留在另一个桌子中.所以Ignatius至少需要2个桌子. 思路: 并查集模板题. #include<iostream> using namespace std; ]; int find(in…

HDU 1814 Peaceful Commission / HIT 1917 Peaceful Commission /CJOJ 1288 和平委员会(2-sat模板题) Description The Public Peace Commission should be legislated in Parliament of The Democratic Republic of Byteland according to The Very Important Law. Unfortunatel…

并查集讲解和模板 有一个博文对此分析的很透彻,附链接 为避免原链接失效,现摘录如下: 为了解释并查集的原理,我将举一个更有爱的例子. 话说江湖上散落着各式各样的大侠,有上千个之多.他们没有什么正当职业,整天背着剑在外面走来走去,碰到和自己不是一路人的,就免不了要打一架.但大侠们有一个优点就是讲义气,绝对不打自己的朋友.而且他们信奉"朋友的朋友就是我的朋友",只要是能通过朋友关系串联起来的,不管拐了多少个弯,都认为是自己人.这样一来,江湖上就形成了一个一个的帮派,通过两两之间的朋友关系串…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3038 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description TT and FF are ... friends. Uh... very very good friends -________-b FF is a bad boy, he is always…

转载:http://www.cnblogs.com/hxer/p/5675149.html 题意:有一个长度为n(n < 5e4)的字符串,Q(Q<=2e5)次操作:操作分为:在末尾插入一个字符ch和查询不同子串出现次数不小于K的数量: 思路1:SAM在线求解: 对于每次找到将一个字符x插入到SAM之后,我们知道pre[p]所含有的Tx的后缀字符串数目为step[pre[np]]个,那么只需要每次插入之后更新下这些字符串出现的次数cnt即可: 由于Right(fa)与Right(r)没有交集(…

畅通工程 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 62539    Accepted Submission(s): 33472 Problem Description 某省调查城镇交通状况,得到现有城镇道路统计表,表中列出了每条道路直接连通的城镇.省政府"畅通工程"的目标是使全省任何两个城镇间都可以实现交通(但不一定有直…