Find them, Catch them

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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 question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)  Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:  1. D [a] [b]  where [a] and [b] are the numbers of two criminals, and they belong to different gangs.  2. A [a] [b]  where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang. 

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.

Output

For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same gang.", "In different gangs." and "Not sure yet."

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.
 #include<stdio.h>
#include<set>
#include<map>
#include<algorithm>
#include<string.h>
const int M = 1e5 + ;
int T ;
int n , m ;
int f[M] ;
int mm[M] ;
int Union (int x)
{
return f[x] == x ? x : f[x] = Union (f[x]) ;
} int main ()
{
//freopen ("a.txt" , "r" , stdin ) ;
scanf ("%d" , &T) ;
while (T --) {
scanf ("%d%d" , &n , &m);
memset (mm , , sizeof(mm)) ;
for (int i = ; i <= n ; i ++) f[i] = i ;
char s[] ;
int _u , _v , u , v ;
int x , y ;
int _x , _y ;
while (m --) {
scanf ("%s" , s) ;
if (s[] == 'D') {
scanf ("%d%d" , &u , &v ) ;
if (mm[u] < mm[v]) std::swap (u , v) ;
if (mm[u] && mm[v]) {
_u = mm[u] ; _v = mm[v] ;
x = Union (_u) ; y = Union (v) ;
f[y] = x ;
x = Union (_v) ; y = Union (u) ;
f[y] = x ;
}
else if (mm[u] && mm[v] == ) {
_u = mm[u] ;
x = Union (_u) ; y = Union (v) ;
f[y] = x ;
mm[v] = u ;
}
else if (mm[u] == ) {
mm[u] = v ;
mm[v] = u ;
}
}
else if (s[] == 'A') {
scanf ("%d%d" , &u , &v) ;
if (mm[u] < mm[v]) std::swap (u , v) ;
if ( !mm[u] || !mm[v]) puts ("Not sure yet.") ;
else {
x = Union (u) ; y = Union (v) ;
if (x == y) puts ("In the same gang.") ;
else {
_x = Union (mm[u]) ;
if (_x != y) puts ("Not sure yet.") ;
else puts ("In different gangs.") ;
}
}
}
}
}
return ;
}

poj.1703.Find them, Catch them(并查集)的更多相关文章

  1. POJ 2236 Wireless Network ||POJ 1703 Find them, Catch them 并查集

    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 ...

  2. POJ 1703 Find them, catch them (并查集)

    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 ...

  3. POJ 1703 Find them, Catch them 并查集的应用

    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 ...

  4. POJ 1703 Find them, Catch them(并查集高级应用)

    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 ...

  5. POJ 1703 Find them, Catch them 并查集,还是有点不理解

    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/freezhanacmor ...

  6. [并查集] POJ 1703 Find them, Catch them

    Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 43132   Accepted: ...

  7. POJ 1703 Find them, Catch them(种类并查集)

    Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 41463   Accepted: ...

  8. hdu - 1829 A Bug's Life (并查集)&&poj - 2492 A Bug's Life && poj 1703 Find them, Catch them

    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 ...

  9. POJ 1703 Find them, Catch them (数据结构-并查集)

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

随机推荐

  1. 基本概率分布Basic Concept of Probability Distributions 2: Poisson Distribution

    PDF version PMF A discrete random variable $X$ is said to have a Poisson distribution with parameter ...

  2. COGS 577 蝗灾

    传送门 时间限制:2 s 内存限制:128 MB DESCRIPTION C国国土辽阔,地大物博......但是最近却在闹蝗灾..... 我们可以把C国国土当成一个W×W的矩阵,你会收到一些诸如(X, ...

  3. ZooKeeper配置管理文件

    最近在工作中,为了完善公司集群服务的架构,提高可用性,降低运维成本,因此开始学习ZooKeeper.    至于什么是ZooKeeper?它能做什么?如何安装ZooKeeper?我就不一一介绍了,类似 ...

  4. Android中如何像 360 一样优雅的杀死后台Service而不启动

    http://my.oschina.net/mopidick/blog/277813 目录[-] 一.已知的 kill 后台应用程序的方法 方法: kill -9 pid 二.终极方法,杀死后台ser ...

  5. HD1814Peaceful Commission(模板题)

    题目链接 题意: 和平委员会 根据宪法,Byteland民主共和国的公众和平委员会应该在国会中通过立法程序来创立. 不幸的是,由于某些党派代表之间的不和睦而使得这件事存在障碍. 此委员会必须满足下列条 ...

  6. php 如何造一个简短原始的数据库类用来增加工作效率

    class DBDA{ public $host="localhost"; public $uid="root"; public $pwd="123& ...

  7. Canvas绘画功能(待补充)

    由于项目的前端需要用户手绘输入,所以我们利用Canvas控件做绘画面板,并且实现了许多功能,包括手绘笔画,清空画板,上传手绘图,下载手绘图,记录用户笔画,上传背景图.以后有时间都写到这篇博客中,今天晚 ...

  8. MSDeploy 同步时不删除原有文件

    在 jenkins里  Execute Windows batch command "C:\Program Files (x86)\IIS\Microsoft Web Deploy V3\m ...

  9. C# Get/Post 模拟提交

    public static string GetPage(string url, string encoding) { PublicVariables.NetworkConnection = fals ...

  10. iOS / Android 移动设备中的 Touch Icons

    上次转载了一篇<将你的网站打造成一个iOS Web App>,但偶然发现这篇文章的内容有些是错误的——准确来说也不是错误,只是不适合自半年前来的情况了(也可以说是iOS7 之后的时间)—— ...