不想看模板,想直接看题目的请戳下面目录: 目录: HDU 1213 How Many Tables[传送门] HDU 1232 畅通工程 [传送门] POJ 2236 Wireless Network [传送门] POJ 1703 Find them, Catch them [传送门] 先上模板: #define MAXN 根据编号需要 int per[MAXN],rank[MAXN]; void init(int n) { int i; ;i<=n;i++) { per[i]=i;rank[i…

定义&&概念: 啥是并查集,就是将所有有相关性的元素放在一个集合里面,整体形成一个树型结构,它支持合并操作,但却不支持删除操作 实现步骤:(1)初始化,将所有节点的父亲节点都设置为自己,例如pre[1]=1(2)合并,将一个元素或者一集合(两者间有联系)合并到另外一个集合(元素)里面,谁是谁的父亲节点不需要过多在意,视题意而定.(3)查找,在合并时需要运用到查找操作,即查找该元素的父节点,尽量使用路径压缩,可以使并查集更加高效,一旦使用了路径压缩,查询时就会将该查询元素到父亲的边改为直接连…

PS:做到第四题才发现 2,3题的路径压缩等于没写 How Many Tables Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14542    Accepted Submission(s): 7132 Problem Description Today is Ignatius' birthday. He invites a lot…

一开始以为两道题是一样的,POJ的过了直接用相同代码把HDU的交了,结果就悲剧了.最后发现HDU的没有考虑入度不能大于一. 题意:用树的定义来 判断吧,无环,n个结点最多有n-1条边,不然就会有环.只有一个入度为0的结点,不存在入度大于1的结点. 思路:并查集. AC代码: #include<stdio.h> #include<string.h> #define N 100005 int in[N],pre[N],a,b,c[N]; void init()//初始化 { for(i…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=1213 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82830#problem/C 代码: #include<stdio.h> #include<queue> #include<stack> #include<string.h> using namespace std; #define maxn 100…

题目描述 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, and all the friends do not want to stay with s…

#include<iostream> using namespace std; ; int p[N]; int find(int x) { if(p[x]!=x) p[x]=find(p[x]); return p[x]; } int main() { int t; int n,m; cin>>t; while(t--) { cin>>n>>m; ;i<=n;i++) p[i]=i; while(m--) { int a,b; cin>>a…

并查集的介绍可以看下https://www.cnblogs.com/jkzr/p/10290488.html A - Wireless Network POJ - 2236 An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected afters…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1213 Problem Description 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 fri…

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,…