tree  Accepts: 143  Submissions: 807  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 有一个树(nn个点, n-1n−1条边的联通图),点标号从11~nn,树的边权是00或11.求离每个点最近的点个数(包括自己). 输入描述 第一行一个数字TT,表示TT组数据. 对于每组数据,第一行是一个nn,表示点个数,接下来n-1n−1,每行三个…

  The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question…

D. Dividing Kingdom II   Long time ago, there was a great kingdom and it was being ruled by The Great Arya and Pari The Great. These two had some problems about the numbers they like, so they decided to divide the great kingdom between themselves. Th…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3018 题目大意:有n个点,m条边,人们希望走完所有的路,且每条道路只能走一遍.至少要将人们分成几组. 解题思路:先用并查集求出所有的连通块,然后判断每个连通块内点的度数,如果有奇数点则需要的组数ans+=奇数点/2:反之,所需组数ans+=1.注意:如果遇到孤立点即度数为0的点则不用算进去,因为没有跟他相连的边. 代码: #include<iostream> #include<cstdio&…

目录 认识并查集 并查集解析 基本思想 如何查看a,b是否在一个集合? a,b合并,究竟是a的祖先合并在b的祖先上,还是b的祖先合并在a上? 其他路径压缩? 代码实现 结语 @(文章目录) 认识并查集 对于并查集(不相交集合),很多人会感到很陌生,没听过或者不是特别了解.实际上并查集是一种挺高效的数据结构.实现简单,只是所有元素统一遵从一个规律所以让办事情的效率高效起来. 对于定意义,百科上这么定义的: 并查集,在一些有N个元素的集合应用问题中,我们通常是在开始时让每个元素构成一个单元素的集合,…

C. Edgy Trees time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a tree (a connected undirected graph without cycles) of nn vertices. Each of the n−1n−1 edges of the tree is col…

Description Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements. Mr Wang selected a room big enough to hold the boys. The boy w…

Description Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements. Mr Wang selected a room big enough to hold the boys. The boy w…

Conscription Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14661   Accepted: 5102 Description Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to b…

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 friends know each other, and all the friends do not want to stay…

Description Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others. In t…

hdu1233 还是畅通工程 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21371    Accepted Submission(s): 9515 Problem Description 某省调查乡村交通状况,得到的统计表中列出了任意两村庄间的距离.省政府"畅通工程"的目标是使全省任何两个村庄间都可以实现公路交通(但不一…

More is better Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)Total Submission(s): 25168    Accepted Submission(s): 9045 Problem Description Mr Wang wants some boys to help him with a project. Because the project…

战争中保持各个城市间的连通性非常重要.本题要求你编写一个报警程序,当失去一个城市导致国家被分裂为多个无法连通的区域时,就发出红色警报.注意:若该国本来就不完全连通,是分裂的k个区域,而失去一个城市并不改变其他城市之间的连通性,则不要发出警报. 输入格式: 输入在第一行给出两个整数N(0 < N <=500)和M(<=5000),分别为城市个数(于是默认城市从0到N-1编号)和连接两城市的通路条数.随后M行,每行给出一条通路所连接的两个城市的编号,其间以1个空格分隔.在城市信息之后给出被攻…

Description 抗日战争时期,冀中平原的地道战曾发挥重要作用. 地道的多个站点间有通道连接,形成了庞大的网络.但也有隐患,当敌人发现了某个站点后,其它站点间可能因此会失去联系. 我们来定义一个危险系数DF(x,y): 对于两个站点x和y (x != y), 如果能找到一个站点z,当z被敌人破坏后,x和y不连通,那么我们称z为关于x,y的关键点.相应的,对于任意一对站点x和y,危险系数DF(x,y)就表示为这两点之间的关键点个数. 本题的任务是:已知网络结构,求两站点之间的危险系数. In…

Description Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others. In t…

/* 模拟二分图:每个点作为一条边,连接的是一列和一行(抽象成一个点,列在左,行在右) 由题意得 a-b相连,a-c相连,b-d相连,那么d-c就不用再相连了 等价于把二分图变成联通的需要再加多少边 用并查集可以解决 */ #include<bits/stdc++.h> using namespace std; #define maxn 400005 int F[maxn],n,m,q; int find(int x){ return F[x]==x?x:F[x]=find(F[x]); }…

Description One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in the room but didn’t find the hamster. He tried to use some cheese to trap the hamster. He put the cheese trap in his room and waited for three da…

Dragon Balls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4384    Accepted Submission(s): 1673 Problem Description Five hundred years later, the number of dragon balls will increase unexpecte…

Person p1 = new Person("张三", new BigDecimal("10.0"));Person p2 = new Person("李四", new BigDecimal("30.0"));Person p3 = new Person("王五", new BigDecimal("40.0"));Person p4 = new Person("赵六"…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2545 思路:dist[u]表示节点u到根节点的距离,然后在查找的时候更新即可,最后判断dist[u],dist[v]的大小即可. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std;…

The Suspects Time Limit: 1000MS   Memory Limit: 20000K Total Submissions: 28164   Accepted: 13718 Description Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. T…

http://www.lydsy.com/JudgeOnline/problem.php?id=1015 题意: 思路:好题啊!!! 这道题目需要离线处理,先把所有要删的点给保存下来,然后逆序加点,这样就把原来的删点变为了加点,加点的话计算连通块就方便的多,具体参见代码. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<sstream&…

代码: #include<cstdio> #include<cstring> using namespace std; int n,m; int father[1005]; int Find(int a) { int r=a; while(father[a]!=a) { a=father[a]; } father[r]=a; return a; } void Union(int a,int b) { a=Find(a); b=Find(b); if(a!=b) father[b]=…

Description It’s an interesting experience to move from ICPC to work, end my college life and start a brand new journey in company. As is known to all, every stuff in a company has a title, everyone except the boss has a direct leader, and all the re…

Description ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the…

Conquer a New Region Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 657 Accepted Submission(s): 179 Problem Description The wheel of the history rolling forward, our king conquered a new region i…

并查集专题训练地址,注册登录了才能看到题目 并查集是一个树形的数据结构,  可以用来处理集合的问题, 也可以用来维护动态连通性,或者元素之间关系的传递(关系必须具有传递性才能有并查集来维护,因为并查集有压缩路径). 用并查集来维护元素之间关系的传递, 那么元素与元素之间就有一个权值(带权并查集),那么当路径压缩时,我们需要根据关系的传递性来维护 当父元素变动时,关系的变动. poj1182食物链 给定一些元素之间的关系, 问我们有多少是错误的. 当碰到一个错误时,并不会用它来更新并查集(错误的怎…

一.并查集的概念:     首先,为了引出并查集,先介绍几个概念:     1.等价关系(Equivalent Relation)     自反性.对称性.传递性.     如果a和b存在等价关系,记为a~b.     2.等价类:     一个元素a(a属于S)的等价类是S的一个子集,它包含所有与a有关系的元素.注意,等价类形成对S的一个划分:S的每一个成员恰好互斥地出现在一个等价类中.为了确定是否a~b,我们仅需验证a和b是否属于同一个等价类即可.     3.并查集:     即为等价类,…

题意:有N名来自两个帮派的坏蛋,已知一些坏蛋两两不属于同一帮派,求判断给定两个坏蛋是否属于同一帮派. 思路: 解法一: 编号划分 定义并查集为:并查集里的元素i-x表示i属于帮派x,同一个并查集的元素同时成立 可见所有元素个数为2 * N,如果i表示属于帮派A,那么i + N表示属于帮派B,每次输入两个人不在同一帮派的时候,就合并他们分属两个帮派的元素. #include <iostream> using namespace std; #define MAX_N 100000 * 2 + 16…