hdu2473并查集的删除操作建立虚点,删除它就断掉了它在原图中的所有关系,而成为独立节点,而且它只能被删除一次,而且删除之后还能进行操作,采用映射(虚点)的方法,建立虚点并把删除之后的操作挪到虚点上来.啊,初始化确实有问题,有可能多次删除的,所以要按操作的数量来,比如全部删除等等数据,哭了 #include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<…

欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - HDU2473 题意概括 一堆点. 要你支持合并两组点.分离某组点中的一个,这两种操作. 点数<=100000,操作数<=1000000 题解 删除点不难,只需要把之前那个点不删除,然后再建立一个新的点就可以了.具体直接看代码应该就懂了. 代码 #include <cstring> #include <cstdio> #include <algorithm> #incl…

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…

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…

题解: 代码: #include<bits/stdc++.h> using namespace std; #define ll long long ; int fa[maxn],id,vis[maxn]; int find(int x) { if (x!=fa[x]) fa[x]=find(fa[x]); return(fa[x]); } void merge(int x,int y) { int fx=find(x),fy=find(y); if (fx!=fy) fa[fx]=fy; }…

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 whether…

传送门 题意: 每次合并两份邮件,或者将某一份邮件独立出来,问最后有多少个邮件集合. 分析: 考虑初始化每个节点的祖先为一个虚父节点(i + n),虚父节点指向它自己.这样可以进行正常的合并操作. 而在删除时,将该节点的祖先置为一个更大的数(++anc_cnt),并让该anc_cnt指向它自己,这样不会影响与自身节点相连的所有关系.最后统计有多少个不同的祖先即可. code #include<bits/stdc++.h> using namespace std; const int N = 1…

原题链接:https://vjudge.net/problem/11782/origin 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…