链接:

https://vjudge.net/problem/Gym-100923H

题意:

Oberyn Martell and Gregor Clegane are dueling in a trial by combat. The fight is extremely important, as the life of Tyrion Lannister is on the line. Oberyn and Gregor are measuring their skill in combat the only way the two best fighters in Westeros can, a match of Starcraft. The one who supervises the match in none other than Por Costel the pig.

Oberyn and Gregor are both playing the Terrans, and they confront each other in the middle of the map, each with an army of Marines. Unfortunately, pigs cannot distinguish colors that well, that is why Por Costel can't figure out which marine belongs to which player. All he sees is marines in the middle of the map and, from time to time, two marines shooting each other. Moreover, it might be the case that Por Costel's imagination will play tricks on him and he will sometimes think two marines are shooting each other even though they are not.

People are starting to question whether Por Costel is the right person for this important job. It is our mission to remove those doubts. You will be given Por Costel's observations. An observation consists in the fact that Por Costel sees that marine and marine are shooting each other. We know that marines in the same team (Oberyn's or Gregor's) can never shoot each other. Your task is to give a verdict for each observation, saying if it is right or not.

An observation of Por Costel's is considered correct if, considering this observation true and considering all the correct observations up to this point true, there is a way to split the marines in "Oberyn's team" and "Gregor's team" such that no two marines from the same team have ever shot each other. Otherwise, the observation is considered incorrect.

"Elia Martell!!! You rushed her! You cheesed her! You killed her SCVs!"

思路:

带权并查集, 维护每个节点与根节点的关系,为0是友军,为1是敌军,

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL; const int MAXN = 1e5+10;
int Fa[MAXN], Level[MAXN];
int n, m; int GetF(int x)
{
if (Fa[x] == x)
return x;
int tmp = Fa[x];
Fa[x] = GetF(Fa[x]);
Level[x] = Level[x]^Level[tmp];
return Fa[x];
}
//1 1 3
//0 1 0
//1 2 3
//0 same 1 diff
int main()
{
// freopen("meciul.in", "r", stdin);
// freopen("meciul.out", "w", stdout);
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--)
{
cin >> n >> m;
for (int i = 1;i <= n;i++)
Fa[i] = i, Level[i] = 0;
int l, r;
while (m--)
{
cin >> l >> r;
int tl = GetF(l);
int tr = GetF(r);
if (tl != tr)
{
Level[tr] = 1^Level[l]^Level[r];
Fa[tr] = tl;
cout << "YES" << endl;
}
else
{
if (Level[l] == Level[r])
cout << "NO" << endl;
else
cout << "YES" << endl;
}
// for (int i = 1;i <= n;i++)
// cout << Level[i] << ' ' ;
// cout << endl;
}
} return 0;
}

Gym-100923H-Por Costel and the Match(带权并查集)的更多相关文章

  1. 【并查集】Gym - 100923H - Por Costel and the Match

    meciul.in / meciul.out Oberyn Martell and Gregor Clegane are dueling in a trial by combat. The fight ...

  2. 【带权并查集】Gym - 100923H - Por Costel and the Match

    裸题. 看之前的模版讲解吧,这里不再赘述了. #include<cstdio> #include<cstring> using namespace std; int fa[10 ...

  3. K - Find them, Catch them POJ - 1703 (带权并查集)

    题目链接: K - Find them, Catch them POJ - 1703 题目大意:警方决定捣毁两大犯罪团伙:龙帮和蛇帮,显然一个帮派至少有一人.该城有N个罪犯,编号从1至N(N<= ...

  4. POJ 1703 Find them, Catch them(带权并查集)

    传送门 Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 42463   Accep ...

  5. [NOIP摸你赛]Hzwer的陨石(带权并查集)

    题目描述: 经过不懈的努力,Hzwer召唤了很多陨石.已知Hzwer的地图上共有n个区域,且一开始的时候第i个陨石掉在了第i个区域.有电力喷射背包的ndsf很自豪,他认为搬陨石很容易,所以他将一些区域 ...

  6. poj1417 带权并查集 + 背包 + 记录路径

    True Liars Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2713   Accepted: 868 Descrip ...

  7. poj1984 带权并查集(向量处理)

    Navigation Nightmare Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 5939   Accepted: 2 ...

  8. 【BZOJ-4690】Never Wait For Weights 带权并查集

    4690: Never Wait for Weights Time Limit: 15 Sec  Memory Limit: 256 MBSubmit: 88  Solved: 41[Submit][ ...

  9. hdu3038(带权并查集)

    题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=3038 题意: n表示有一个长度为n的数组, 接下来有m行形如x, y, d的输入, 表示 ...

随机推荐

  1. Jenkins持续集成环境部署

    一.下载Jenkins Jenkins下载地址:https://jenkins.io/download/ 这里我们下载的是jenkins.war 二.启动Jenkins 在Linux下启动Jenkin ...

  2. simple vimrc for python

    "显示行数,设置软回车和缩进还有语法set numberset expandtabset tabstop=8set shiftwidth=4set softtabstop=4set auto ...

  3. python学习之不要在列表迭代的时候进行增删操作

    注意:列表不能在for循环时使用remove方法 li = [11,22,33,44] for i in li : li.remove(i) print (li) #输出 [22, 44] ​ for ...

  4. win10序列号 2019年10月测试

    win10序列号 N3415-266GF-AH13H-WA3UE-5HBT4 win10序列号 NPK3G-4Q81M-X4A61-D553L-NV68D win10序列号 N617H-84K11-6 ...

  5. 交换机安全学习笔记 第六章 IPV4 ARP攻击

    ARP欺骗攻击 常用工具:  dsniff(Linux/windows).ettercap(Linux/windows).cain(仅windows). ARP欺骗攻击的目的是嗅探发往某主机的所有IP ...

  6. Java学习开发第三阶段总结

    第三阶段的学习总结: 在这次学习我学习了面向对象和封装的知识. ①类的定义 package day01; public class student { //成员变量 String name; //姓名 ...

  7. CentOS7安装配置MariaDB(mysql)数据主从同步

    CentOS7安装MariaDB并配置主从同步 环境声明: 防火墙firewalld及SElinux均为关闭状态 主库节点:192.168.0.63 从库节点:192.168.0.64 配置主库节点: ...

  8. CDH的ntp时间同步

    云服务器: ntpq -p ntpdate -u 10.52.255.1  #手动同步 自建NTP服务器: https://www.cnblogs.com/yinzhengjie/p/9480665. ...

  9. 洛谷 P4779 单源最短路径(标准版) 题解

    题面 这道题就是标准的堆优化dijkstra: 注意堆优化的dijkstra在出队时判断vis,而不是在更新时判断vis #include <bits/stdc++.h> using na ...

  10. 关于float的小奥秘

    一. float 存储方式 1.1. float 占四个字节 1.2. 浮点数构成 1.2.1. 无论是单精度还是双精度在存储中都分为三个部分: <1>. 符号位(Sign) : 0代表正 ...