【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 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!"
Input
The input file meciul.in will contain, on its first line, the number of tests (
). A test has the following structure: the first line contains integers
(
) and
(
) and the next
lines each contain a pair of integers
and
(
) describing an observation of Por Costel's.
Output
The output file meciul.out will contain one line for each of Por Costel's observations, on each test. The line will contain "YES" if the observation is correct and "NO" otherwise. It is not necessary to leave extra spacing between the outputs of different test cases.
Example
1
3 3
1 2
2 3
1 3
YES
YES
NO
给你一些信息(x,y),告诉你x和y是敌人,让你根据某条信息之前的合法信息,判断当前信息是否合法。
并查集。记录x的敌人集合为enemy[x],只需要存一个代表元即可。
然后如果某条信息合法的话,就将x和enemy[y]合并,y和enemy[x]合并即可。
很像当年入门的时候一道并查集经典题“团伙”。
#include<cstdio>
#include<cstring>
using namespace std;
int fa[100010],__rank[100010],enemy[100010];
int T,n,m;
int findroot(int x)
{
return x==fa[x] ? x : fa[x]=findroot(fa[x]);
}
int Union(int U,int V)
{
if(__rank[U]<__rank[V])
fa[U]=V;
else
{
fa[V]=U;
if(__rank[U]==__rank[V])
++__rank[U];
}
}
int main()
{
//freopen("h.in","r",stdin);
freopen("meciul.in","r",stdin);
freopen("meciul.out","w",stdout);
int x,y;
scanf("%d",&T);
for(;T;--T)
{
memset(__rank,0,sizeof(__rank));
memset(enemy,0,sizeof(enemy));
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i)
fa[i]=i;
for(int i=1;i<=m;++i)
{
scanf("%d%d",&x,&y);
int f1=findroot(x),f2=findroot(y);
if(f1==f2)
puts("NO");
else
{
if(!enemy[x])
enemy[x]=y;
else
Union(f2,findroot(enemy[x]));
if(!enemy[y])
enemy[y]=x;
else
Union(f1,findroot(enemy[y]));
puts("YES");
}
}
}
return 0;
}
【并查集】Gym - 100923H - Por Costel and the Match的更多相关文章
- 【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 ...
- 【Heap-dijkstra】Gym - 100923B - Por Costel and the Algorithm
algoritm.in / algoritm.out Even though he isn't a student of computer science, Por Costel the pig ha ...
- 【找规律】Gym - 100923L - Por Costel and the Semipalindromes
semipal.in / semipal.out Por Costel the pig, our programmer in-training, has recently returned from ...
- 【分块打表】Gym - 100923K - Por Costel and the Firecracker
semipal.in / semipal.out Por Costel the pig, our programmer in-training, has recently returned from ...
- 【数形结合】Gym - 100923I - Por Costel and the Pairs
perechi3.in / perechi3.out We don't know how Por Costel the pig arrived at FMI's dance party. All we ...
- 【动态规划】Gym - 100923A - Por Costel and Azerah
azerah.in / azerah.out Por Costel the Pig has received a royal invitation to the palace of the Egg-E ...
- Gym100923H Por Costel and the Match
题目链接:http://codeforces.com/gym/100923/problem/H 分析:并查集,用enemy储存x的敌人,用weight储存权重决定根节点 需用scanf和puts输入输 ...
- Gym-100923H-Por Costel and the Match(带权并查集)
链接: https://vjudge.net/problem/Gym-100923H 题意: Oberyn Martell and Gregor Clegane are dueling in a tr ...
- Codeforces Gym 100463E Spies 并查集
Spies Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100463/attachments Desc ...
随机推荐
- mybatis的注解功能
一.mybatis 简单注解 关键注解词 : @Insert : 插入sql , 和xml insert sql语法完全一样 @Select : 查询sql, 和xml select sql语法完全一 ...
- mysql修改表中某个字段的默认值
Mysql中用SQL增加.删除字段,修改字段名.字段类型.注释,调整字段顺序总结 在网站重构中,通常会进行数据结构的修改,所以添加,删除,增加mysql表的字段是难免的,有时为了方便,还会增加修改 ...
- bzoj 4004 [JLOI2015]装备购买 拟阵+线性基
[JLOI2015]装备购买 Time Limit: 20 Sec Memory Limit: 128 MBSubmit: 1820 Solved: 547[Submit][Status][Dis ...
- keydown
<!DOCTYPE HTML><html><head> <meta charset="utf-8"> <title>无标 ...
- 使用bcrypt进行加密的简单实现
Bcrypt百度百科: bcrypt,是一个跨平台的文件加密工具.由它加密的文件可在所有支持的操作系统和处理器上进行转移.它的口令必须是8至56个字符,并将在内部被转化为448位的密钥. 除了对您的数 ...
- 转:Linux 目录结构和常用命令
转自:http://www.cnblogs.com/JCSU/articles/2770249.html仅为学习参考之用 一.Linux目录结构 你想知道为什么某些程序位于/bin下,或者/sbin, ...
- shell 灵活设置定时任务
#!/bin/bash step=30 #间隔的秒数,不能大于60 for (( i = 0; i < 60; i=(i+step) )); do curl #调用链接 sleep $step ...
- GitLab版本管理【转】
转自:http://www.cnblogs.com/wintersun/p/3930900.html GitLab是利用 Ruby on Rails 一个开源的版本管理系统,实现一个自托管的Git项目 ...
- Linux下安装php环境并且配置Nginx支持php-fpm模块[www]
Linux下安装php环境并且配置Nginx支持php-fpm模块 http://www.cnblogs.com/freeweb/p/5425554.html 5分钟搭建 nginx +php --- ...
- Guava Cache相关
官方:http://ifeve.com/google-guava-cachesexplained/ 理解:https://segmentfault.com/a/1190000007300118 项目中 ...