直秒并查集.这题的难点就在于怎么删点.如果要删的是叶节点,那还好,直接刨掉即可 如果是中间节点甚至是根节点,那就不好办了..... solution: 对于独立一个点,我可以用邻接表模拟,然后用并查集维护联通,删点就是普通删点,但是实现难度高,复杂度大,算了,还是想正解吧 正解:对于一个删了的点,我们打个标记,然后开一个新点表示这个点(居然有点像主席树233.....) 于是,动态开点并查集诞生了(口胡) 具体操作在代码里解释 代码: #include<bits/stdc++.h> using…

Junk-Mail Filter Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9135    Accepted Submission(s): 2885 Problem Description Recognizing junk mails is a tough task. The method used here consists o…

给n组操作,每组操作形式为x y p. 当p为1时,如果第x变量和第y个变量可以相等,则输出YES,并限制他们相等:否则输出NO,并忽略此次操作. 当p为0时,如果第x变量和第y个变量可以不相等,则输出YES,并限制他们不相等 :否则输出NO,并忽略此次操作. Input 输入一个数n表示操作的次数(n<=1*10^5) 接下来n行每行三个数x,y,p(x,y<=1*10^8,p=0 or 1) Output 对于n行操作,分别输出n行YES或者NO Input示例 3 1 2 1 1 3 1…

Junk-Mail Filter Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6276    Accepted Submission(s): 1981 Problem Description Recognizing junk mails is a tough task. The method used here consists o…

Colored Sticks DescriptionYou are given a bunch of wooden sticks. Each endpoint of each stick is colored with some color. Is it possible to align the sticks in a straight line such that the colors of the endpoints that touch are of the same color?Inp…

Junk-Mail Filter Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7515    Accepted Submission(s): 2368 Problem Description Recognizing junk mails is a tough task. The method used here consists o…

Problem Description Recognizing junk mails is a tough task. The method used here consists of two steps:1) Extract the common characteristics from the incoming email.2) Use a filter matching the set of common characteristics extracted to determine whe…

HDU2473 Junk-Mail Filter Problem Description Recognizing junk mails is a tough task. The method used here consists of two steps: Extract the common characteristics from the incoming email. Use a filter matching the set of common characteristics extra…

题目链接 没删除调试输出,原地炸裂,\(80\)->\(0\).如果你要问剩下的\(20\)呢?答:数组开小了. 这题正向删边判连通性是很不好做的,因为我们并不会并查集的逆操作.于是可以考虑把断边改成逆向连边,某个猴子什么时候和\(1\)号猴子变成连通的,这就是他掉下去的时间,如果本来就与\(1\)号猴子连通,那么它就永远不会掉下去.我用了两个并查集,一个维护连通性,一个维护答案,这两个并查集前面的操作几乎是一模一样的,同时连边,到最后一步也就是某个猴子与\(1\)号猴子变为连通时,前者正常连边…

Junk-Mail Filter Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8640    Accepted Submission(s): 2735 Problem Description Recognizing junk mails is a tough task. The method used here consists…