题目: 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 questio…

POJ 2236 Wireless Network http://poj.org/problem?id=2236 题目大意: 给你N台损坏的电脑坐标,这些电脑只能与不超过距离d的电脑通信,但如果x和y均能C通信,则x和y可以通信.现在给出若干个操作, O p 代表修复编号为p的电脑 S p q代表询问p和q是不是能通信. 思路: 并查集即可.. 如果修复了一台电脑,则把与它相连距离不超过d的且修复了的放在一个集合里面. #include<cstdio> #include<cstring&…

Find them, Catch them Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1703 Description The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs…

题目:Find them,Catch them 刚开始以为是最基本的并查集,无限超时. 这个特殊之处,就是可能有多个集合. 比如输入D 1 2  D 3 4 D 5 6...这就至少有3个集合了.并且任意2个集合之间成员的敌我关系不明. 这里每个集合里面的成员关系要记录,他们在一个集合里,只是因为他们关系明确,敌人或者朋友. 千万不要简单的认为成朋友在一个集合,敌人在另外一个集合,分成2个集合.因为题目说只有2个匪帮,很容易进入这个误区. 我们只要记录一个成员和自己父亲的敌我关系和建树就可以了.…

题意:城市中有两个帮派,输入中有情报和询问.情报会告知哪两个人是对立帮派中的人.询问会问具体某两个人的关系. 思路:并查集的应用.首先,将每一个情报中的两人加入并查集,在询问时先判断一下两人是否在一个集合中,如果是,则表明两个人的关系已知. 本题还需要判断出两人是不是在同一帮派,这里用一个relation数组来维护每个人与根结点的关系. 0表示该人与根节点为同一个帮派. 1表示该人与根节点为对立帮派. 并查集有两个基本操作:find——查找根节点, merge——合并两个集合. 在合并两个集合时…

手动博客搬家:本文发表于20170805 21:25:49, 原地址https://blog.csdn.net/suncongbo/article/details/76735893 URL: http://poj.org/problem?id=1703 题目大意:本题即很经典的"龙帮虎帮"问题. 有n个元素(n<=1e5),分布在两个不同的集合里. 现在有M个语句(m<=1e5),每个语句共两种:(1) 给定某两个元素在不同的集合中.(2) 询问两个元素是否在同一集合中.对…

题目不难理解,A判断2人是否属于同一帮派,D确认两人属于不同帮派.于是需要一个数组r[]来判断父亲节点和子节点的关系.具体思路可参考http://blog.csdn.net/freezhanacmore/article/details/8774033 #include <iostream> #include <cstring> #include <cstdio> using namespace std; const int maxn=100005; int f[maxn…

The Suspects Time Limit: 1000MS   Memory Limit: 20000K Total Submissions: 24134   Accepted: 11787 Description Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. T…

Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 43132   Accepted: 13257 Description 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 Drago…

Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 41463   Accepted: 12753 Description 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 Drago…

http://acm.hdu.edu.cn/showproblem.php?pid=1829 http://poj.org/problem?id=2492 臭虫有两种性别,并且只有异性相吸,给定n条臭虫(编号1-n)和m对关系,判断是否是出现同性恋的情况. 这题跟食物链的题类似,这里只有两种关系,关系是同性或者异性. 对于每只动物创建两个元素  同性或异性,并用这 2×n个元素建立并查集. 首先判断输入的x和y是否是一个组的.是就flag=1. 然后,如果x和y是一对,那么合并x和y+n,x+n…

Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31102   Accepted: 9583 Description 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…

题目地址:http://poj.org/problem?id=1703 Sample Input 1 5 5 A 1 2 D 1 2 A 1 2 D 2 4 A 1 4 Sample Output Not sure yet. In different gangs. In the same gang. 题目解读:T组数据,每组输入n m,n代表有n个人, m代表有m次操作.操作A x y:表示询问x y的关系是怎样的? 三中情况:不确定,属于同一个集合, 不属于同一个集合操作D x y:表示x y…

Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 36176   Accepted: 11090 Description 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 Drago…

传送门 Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 42463   Accepted: 13065 Description 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 D…

题目:http://poj.org/problem?id=1703 题意:一个地方有两个帮派, 每个罪犯只属于其中一个帮派,D 后输入的是两个人属于不同的帮派, A后询问 两个人是否属于 同一个帮派. 用op[]数组记录 不跟自己在一个帮派中的一个人, 然后再输入不跟自己在一个帮派的人的时候,把 这个人跟 op[]数组记录里的那个人联系起来, 这两个人属于一个帮派. #include <iostream> #include <cstdio> #include <cstring…

Description 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…

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…

<题目链接> 题目大意: 已知所有元素要么属于第一个集合,要么属于第二个集合,给出两种操作.第一种是D a b,表示a,b两个元素不在一个集合里面.第二种操作是A a b,表示询问a,b两个元素是否在同一个集合里面.如果不能确定的话打印not sure.否则给出答案. 解题分析: 典型的带权并查集,通过点与根节点的关系来判断它们是否属于同一阵营,而点与根节点的关系又是在状态压缩时,通过它与父亲节点的关系来更新. #include <cstdio> #include <cstr…

Description 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…

  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…

poj(1703) Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26992   Accepted: 8188 Description 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, G…

Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 32073   Accepted: 9890 Description 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…

这题首先不说怎么做,首先要提醒的是..:一定不要做成多组输入,,我WA了一个晚上加上午,,反正我是尝到苦头了,,请诸君千万莫走这条弯路..切记 这题是上一题(Find them and Catch them)的难度更高的版本,如果你没做的话建议先做那个,用并查集来解,分成三种状态,因为要查询关系,直接查无法查,于是以根节点作为中继点,每个节点做一个标记表示与根节点的关系,0表示与根同类,1表示吃根,2表示被根吃,也必须是这三个数,不能使-1,1,0或别的什么的,因为那样无法转化. 在findse…

http://poj.org/problem?id=1733 题意: 给出一个01串,有多次询问,每次回答[l,r]这个区间内1的个数的奇偶性,但是其中有一些回答是错误的,问到第几个回答时与前面的回答是有矛盾的. 思路: 任意一个区间要么是奇要么就是偶.所有就可以用种类并查集来解决. 因为是区间,所以如果要连起来的话,每个区间的左端点需要减1. 因为n很大但是询问少,所有需要离散化处理. 下面的代码中,1表示奇,0表示偶. #include<iostream> #include<cstd…

题目链接:http://poj.org/problem?id=1984 Time Limit: 2000MS Memory Limit: 30000K Case Time Limit: 1000MS Description Farmer John's pastoral neighborhood has N farms (2 <= N <= 40,000), usually numbered/labeled 1..N. A series of M (1 <= M < 40,000)…

题目链接:http://poj.org/problem?id=1417 Time Limit: 1000MS Memory Limit: 10000K Description After having drifted about in a small boat for a couple of days, Akira Crusoe Maeda was finally cast ashore on a foggy island. Though he was exhausted and despair…

题目链接: http://poj.org/problem?id=2524 Ubiquitous Religions Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 39369   Accepted: 18782 Description There are so many different religions in the world today that it is difficult to keep track of…

题意:有n只虫子,每次给出一对互为异性的虫子的编号,输出是否存在冲突. 思路:用并查集,每次输入一对虫子后就先判定一下.如果两者父亲相同,则说明关系已确定,再看性别是否相同,如果相同则有冲突.否则就将两只虫子并入一个集合. 而性别则是用一个gender数组来维护,每个虫子的gender的值为0或者1. 0表示该虫子的性别与父亲结点的性别相同. 1表示该虫子的性别与父亲结点的性别不同. 这题和poj1703本质上是一样的.1703的题解请点传送门. 另外还有一个疑惑,本题输入量巨大,我用输入加速后…

                                         Find them, Catch them 好久没有做并查集的题,竟然快把并查集忘完了. 题意:大致是有两个监狱,n个犯人,m次操作,每次操作可以是查询也可以是确定两个人是否在同一个监狱里. 思路:其实这题完全可能做出来,结果就是没做出来.理清思路还是很好想的,我们用一个并查集来把能够确定在同一监狱里的人放在一个集合里,用一个diff数组把确定不同监狱的两个犯人分别与对方不同监狱的犯人放在一个集合里,就是这么简单.…