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."

题目大意:有n个人,D a b表示a b位于不同的集合,A a b则代表问a b是否属于同一个集合(在线提问)。

思路:并查集拓展的简单应用,不多讲。就用rela[x]表示x与当前父节点(不一点是根节点)是否相同。

代码(313MS):

 #include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; const int MAXN = ; int fa[MAXN];
bool rela[MAXN]; int get_set(int x) {
if(x == fa[x]) return x;
else {
int ret = get_set(fa[x]);
rela[x] ^= rela[fa[x]];
return fa[x] = ret;
}
} void unionSet(int x, int y) {
int fx = get_set(x);
int fy = get_set(y);
fa[fy] = fx;
rela[fy] = ^ (rela[x] ^ rela[y]);
} void query(int x, int y) {
int fx = get_set(x);
int fy = get_set(y);
if(fx != fy) puts("Not sure yet.");
else if(rela[x] == rela[y]) puts("In the same gang.");
else puts("In different gangs.");
} int main() {
int T, n, m, x, y;
char c;
scanf("%d", &T);
while(T--) {
scanf("%d%d", &n, &m);
for(int i = ; i <= n; ++i)
fa[i] = i, rela[i] = ;
while(m--) {
scanf(" %c%d%d", &c, &x, &y);
if(c == 'D') unionSet(x, y);
else query(x, y);
}
}
}

POJ 1703 Find them, Catch them(并查集拓展)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)

    Find them, Catch them Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I6 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

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

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

  8. 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: ...

  9. 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 ...

随机推荐

  1. php中的变量作用域

    <?php include_once $_SERVER['DOCUMENT_ROOT'].'/includes/db.inc.php'; function totalJokes() { try{ ...

  2. o'Reill的SVG精髓(第二版)学习笔记——第七章

    第七章:路径 所有描述轮廓的数据都放在<path>元素的d属性中(d是data的缩写).路径数据包括单个字符的命令,比如M表示moveto,L表示lineto.接着是该命令的坐标信息. 7 ...

  3. Restrramework源码(包含组件)分析

    1.总体流程分析 rest_framework/view.py 请求通过url分发,触发as_view方法,该方法在ViewSetMixin类下 点进去查看as_view源码说明,可以看到它在正常情况 ...

  4. 与select2有关的知识点总结

    1.多选下拉框设置提示 var datass = [ { id:0, text: '你好' }, { id:1, text: '好久不见' }, { id:2, text: '好想你' } ]; va ...

  5. dual表详解

    dual是一个虚拟表,用来构成select的语法规则,oracle保证dual里面永远只有一条记录.我们可以用它来做很多事情,如下: 1.查看当前用户 SQL> select user from ...

  6. zepto 基础知识(4)

    61.prev prev() 类型:collection prev(selector) 类型:collection 获取对相集合中每一个元素的钱一个兄弟节点,通过选择器来进行过滤 62.prev pr ...

  7. IPC进程间通信---消息队列

    消息队列 消息队列:消息队列是一个存放在内核中的消息链表,每个消息队列由消息队列标识符标识.与管道不同的是消息队 列存放在内核中,只有在内核重启(即操作系统重启)或者显式地删除一个消息队列时,该消息队 ...

  8. 关于parseInt的看法

    ​ 前面在看题目的时候 偶然看到 使用parseInt 来进行整数判断 但是这里的parseInt是错误示范 之后了解了一下 发现这和函数 很有研究 先看看 w3c怎么说这个的 parseInt() ...

  9. less学习一

    Less 是一门 CSS 预处理语言,它扩展了 CSS 语言,增加了变量.Mixin.函数等特性,使 CSS 更易维护和扩展. Less 可以运行在 Node 或浏览器端. less文件只有被编译后才 ...

  10. JavaScript实现判断图片是否加载完成的3种方法整理

    JavaScript实现判断图片是否加载完成的3种方法整理 有时候我们在前端开发工作中为了获取图片的信息,需要在图片加载完成后才可以正确的获取到图片的大小尺寸,并且执行相应的回调函数使图片产生某种显示 ...