Find them, Catch them
 
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 32225   Accepted: 9947

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. 题目大意:N 表示共有 N 个帮派;下面有 M 条个询问
有两个帮派,现在告诉你一些互相敌对的点对,然后让你判断某两个点之间的关系,并首先判两个点关系是否确定;
用并查集的思想只要两个点在一个集合里就可以了,直接用并查集然后时候在同一个集合;
AC代码:
 #include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
// 0 代表木有关系,1表示不同类,2代表同类
const int N = 1e5+;
int m;
int f[N],c[N];
int init()
{
for(int i=; i<=m; i++)
{
f[i] = i;
c[i] = ;
}
}
int xfind(int x)
{
if(f[x] == x)
return x;
int t = f[x];
f[x] =xfind(f[x]);
c[x] = (c[x]+c[t])%;
return f[x];
}
int solve(char ch,int x,int y )
{
int fx = xfind(x);
int fy = xfind(y);
if(fx != fy)
{
if(ch == 'A') return ; //返回值0,表示不确定
f[fx] =fy;
if((c[x]+c[y])% ==)
c[fx] = ;
}
else
{
if(ch == 'A')
{
if(c[x] != c[y]) return ; //返回值1,表示在不同的帮派
if(c[x] == c[y]) return ; //返回值2,表示在相同的帮派
}
}
if(ch == 'D') return ;//这种情况不同在考虑的
}
int main( )
{
int T,n,x,y;
char s;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&m,&n);
init(); //初始化并查集
for(int i=; i<n; i++)
{
scanf(" %c %d %d",&s,&x,&y);
int flag = solve(s,x,y);
if(flag == ) printf("Not sure yet.\n");
else if(flag == ) printf("In different gangs.\n");
else if(flag == ) printf("In the same gang.\n");
}
}
return ;
}

poj1703 Find them, Catch them(并查集的应用)的更多相关文章

  1. poj1703 Find them, Catch them 并查集

    poj(1703) Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26992   ...

  2. POJ-1703 Find them, Catch them(并查集&数组记录状态)

    题目: The police office in Tadu City decides to say ends to the chaos, as launch actions to root up th ...

  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. poj1703--Find them, Catch them(并查集应用)

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

  5. POJ1703-Find them, Catch them 并查集构造

                                             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 ...

  6. 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 ...

  7. 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 ...

  8. poj1703_Find them, Catch them_并查集

    Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 42451   Accepted: ...

  9. poj.1703.Find them, Catch them(并查集)

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

  10. 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 ...

随机推荐

  1. Android进阶笔记:Messenger源码详解

    Messenger可以理解为一个是用于发送消息的一个类用法也很多,这里主要分析一下再跨进程的情况下Messenger的实现流程与源码分析.相信结合前面两篇关于aidl解析文章能够更好的对aidl有一个 ...

  2. Java读写文件,中文乱码解决

    读文件:使用new BufferedReader(new InputStreamReader(new FileInputStream(file), "UTF-8")); Strin ...

  3. 网站怎么布局能解决不同浏览器对CSS解析的差异,使用css reset

    很多地方都提到过CSS Reset这个概念,而且细心的朋友会发现,许多大网站的CSS文件中也含有CSS Reset内容. CSS Reset是什么? 在HTML标签在浏览器里有默认的样式,例如 p 标 ...

  4. [转]Sql Server参数化查询之where in和like实现详解

    本文转自;http://www.cnblogs.com/lzrabbit/archive/2012/04/22/2465313.html 文章导读 拼SQL实现where in查询 使用CHARIND ...

  5. Firefox的缓存问题

    使用Firefox调试代码时,JS或CSS文件修改后,怎么刷新都不生效.原来是Firefox缓存的问题.但Firefox没有像IE一样有个每次读取最新的设置. 设置Firefox不缓存页面: 新建标签 ...

  6. java缓存适合使用的情况

    并非所有的情况都适合于使用二级缓存,需要根据具体情况来决定.同时可以针对某一个持久化对象配置其具体的缓存策略. 适合于使用二级缓存的情况: 1.数据不会被第三方修改 一般情况下,会被hibernate ...

  7. Maven Web项目配置Mybatis

    一.添加Mybatis和数据库相关的包 1 pom.xml中添加的包有mybatis,mybatis-spring,druid,MySQL-connector-Java,commons-io,refl ...

  8. 【Linux】apt-get install 怎么阻止弹出框,使用脚本默认自动安装?

    You can do a couple of things for avoiding this. Setting the DEBIAN_FRONTEND variable to noninteract ...

  9. 纯 PHP 代码最好在文件末尾删除 PHP 结束标记

    如果文件内容是纯 PHP 代码,最好在文件末尾删除 PHP 结束标记.这可以避免在 PHP 结束标记之后万一意外加入了空格或者换行符,会导致 PHP 开始输出这些空白,而脚本中此时并无输出的意图. & ...

  10. Swing的GUI组件得到焦点

    Swing的GUI组件如JButtin,JTextArea,JRadioButton,JComboBox等,可以使用requestFocus()方法来获得焦点.