Find them, Catch them
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 42451   Accepted: 13059

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.

Source

 #include <iostream>
#include <cstdio> #define MAX_N 1000000+5 using namespace std; int par[MAX_N];//父节点
int depth[MAX_N];//深度 void init(int n){
for(int i=;i<=n;i++){
par[i]=i;
depth[i]=;
}
}
int find_father(int t){
if(t==par[t]){
return t;
}else{
return par[t]=find_father(par[t]);
//实现了路径压缩
}
}
void unite(int t1,int t2){
int f1=find_father(t1);
int f2=find_father(t2);
if(f1==f2){
return ;
}
if(depth[f1]<depth[f2]){
par[f1]=f2;
}else{
par[f2]=f1;
if(depth[f1]==depth[f2]){
depth[f1]++;
//记录深度
}
}
} bool same(int x,int y){
return find_father(x)==find_father(y);
} int main()
{
int t;
int n,m;
char c;
int a,b;
scanf("%d",&t);
while(t--){
scanf("%d %d",&n,&m);
init(n*);
while(m--){
getchar();
scanf("%c %d %d",&c,&a,&b);
if(c=='D'){
unite(a,b+n);
unite(a+n,b);
}else{
if(same(a,b+n)){
printf("In different gangs.\n");
continue;
}
if(same(a,b)){
printf("In the same gang.\n");
continue;
}else{
printf("Not sure yet.\n");
continue;
}
}
}
}
return ;
}

poj1703_Find them, Catch them_并查集的更多相关文章

  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)

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

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

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

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

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

  9. 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. C/C++实践笔记 004

    转义字符 #define _CRT_SECURE_NO_WARNINGS#include<stdio.h>#include<stdlib.h> void main1() { c ...

  2. 认识和使用NSOperation

    原文链接:http://www.jianshu.com/p/2de9c776f226 NSOperation是OC中多线程技术的一种,是对GCD的OC包装.它包含队列(NSOperationQueue ...

  3. NGUI Tween动画Scale与Transform冲突

    NGUI中我们要同时完成Scale与Transform的效果,会发现动画并不是同我们想的那样运行的. 原因就是Tween Scale与Tween Transform的冲突调用. Tween Scale ...

  4. BZOJ2815: [ZJOI2012]灾难

    传送门 学LCA的时候根本没意识到LCA可以有这么多玩法. 这玩意据说是个高级数据结构(支配树)的弱化版,蒟蒻没学过呀.所以出题人提出一个概念叫灾难树. 我理解的灾难树的意思实际上是属于DAG的一个子 ...

  5. List集合的迭代器方法

    1.后台JAVA代码的实现 //获取所有的支付方式的迭代器 // 获取支付方式 @RequestMapping(value = "get/payed/type", method = ...

  6. 分析DH加密算法,一种适基于密钥一致协议的加密算法。

    DH Diffie-Hellman算法(D-H算法),密钥一致协议.是由公开密钥密码体制的奠基人Diffie和Hellman所提出的一种思想.简单的说就是允许两名用户在公开媒体上交换信息以生成&quo ...

  7. Win10---------专区

    待完善中---------------------------------- -----------------------------------------The End------------- ...

  8. win7使用iis并搭建 图片服务器

    1.打开控制面板 2.程序-卸载程序 3.点击左边的 打开或关闭windows功能 4.如下图所示,找到internet信息服务勾选.顺便把FTP服务器也全部勾选了,后面会用到 5.进入 控制面板 – ...

  9. 盒子模型简单理解(box-sizing)

    普通解析: 概念图示和公式: html结构 <div class="num1"></div> 1.只写 width.height(写背景是为了区分) .nu ...

  10. 关于学习JavaScript 的 高三编程 一些心得

    面对JS 问题来说,很多的细节问题以及 弱类型转换的问题,往往会成为学习js 路上的一个阻碍. 那么问题来了,今天我看到的是  高三 里面的  基本概念的 语法问题. 直奔主题.(还是帖代码先) sw ...