Codeforces Round #345 (Div. 1) E. Clockwork Bomb 并查集
E. Clockwork Bomb
题目连接:
http://www.codeforces.com/contest/650/problem/E
Description
My name is James diGriz, I'm the most clever robber and treasure hunter in the whole galaxy. There are books written about my adventures and songs about my operations, though you were able to catch me up in a pretty awkward moment.
I was able to hide from cameras, outsmart all the guards and pass numerous traps, but when I finally reached the treasure box and opened it, I have accidentally started the clockwork bomb! Luckily, I have met such kind of bombs before and I know that the clockwork mechanism can be stopped by connecting contacts with wires on the control panel of the bomb in a certain manner.
I see n contacts connected by n - 1 wires. Contacts are numbered with integers from 1 to n. Bomb has a security mechanism that ensures the following condition: if there exist k ≥ 2 contacts c1, c2, ..., ck forming a circuit, i. e. there exist k distinct wires between contacts c1 and c2, c2 and c3, ..., ck and c1, then the bomb immediately explodes and my story ends here. In particular, if two contacts are connected by more than one wire they form a circuit of length 2. It is also prohibited to connect a contact with itself.
On the other hand, if I disconnect more than one wire (i. e. at some moment there will be no more than n - 2 wires in the scheme) then the other security check fails and the bomb also explodes. So, the only thing I can do is to unplug some wire and plug it into a new place ensuring the fact that no circuits appear.
I know how I should put the wires in order to stop the clockwork. But my time is running out! Help me get out of this alive: find the sequence of operations each of which consists of unplugging some wire and putting it into another place so that the bomb is defused.
Input
The first line of the input contains n (2 ≤ n ≤ 500 000), the number of contacts.
Each of the following n - 1 lines contains two of integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi) denoting the contacts currently connected by the i-th wire.
The remaining n - 1 lines contain the description of the sought scheme in the same format.
It is guaranteed that the starting and the ending schemes are correct (i. e. do not contain cicuits nor wires connecting contact with itself).
Output
The first line should contain k (k ≥ 0) — the minimum number of moves of unplugging and plugging back some wire required to defuse the bomb.
In each of the following k lines output four integers ai, bi, ci, di meaning that on the i-th step it is neccesary to unplug the wire connecting the contacts ai and bi and plug it to the contacts ci and di. Of course the wire connecting contacts ai and bi should be present in the scheme.
If there is no correct sequence transforming the existing scheme into the sought one, output -1.
Sample Input
3
1 2
2 3
1 3
3 2
Sample Output
1
1 2 1 3
Hint
题意
给你两棵树,你只能操作第一棵树,你每次操作是删除一条边,加一条边
但是都不能构成环,然后问你最少多少步。
题解:
可以强行用LCT做动态最小生成树无脑肝过去就好了。
我推荐rng58的做法:
我们从第一棵树的dfs顺序开始考虑,做到第u个点了,v是u的父亲,如果边(v,u)存在在第二棵树,显然我们就不用动这条边。否则的话,我们将u点连向第二棵树u点的父亲fa[u]就好了。
现在有一个问题,如果第二棵树中u点已经连了父亲fa[u]了,怎么办?
把(u,w)这条边变成(find(u),fa[find(u)])就好了。find是并查集,找到第一个在第一棵树不和自己第二棵树fa[u]相连接的点。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5+7;
vector<int> E[2][maxn];
int fa[2][maxn];
int father[maxn];
int tot=0;
int ans[maxn][4];
void addans(int x,int y,int z,int p)
{
ans[tot][0]=x,ans[tot][1]=y,ans[tot][2]=z,ans[tot][3]=p;
tot++;
}
int fi(int u){
return u != father[u] ? father[u] = fi( father[u] ) : u;
}
bool check(int id,int x,int y)
{
if(x==0||y==0)return false;
if(fa[id][x]==y||fa[id][y]==x)return true;
return false;
}
void dfs(int id,int x,int f)
{
fa[id][x]=f;
for(int i=0;i<E[id][x].size();i++)
{
int v = E[id][x][i];
if(v==f)continue;
dfs(id,v,x);
}
}
void solve(int x,int f)
{
for(int i=0;i<E[0][x].size();i++)
{
int v = E[0][x][i];
if(v==f)continue;
solve(v,x);
if(fa[1][v]!=x&&fa[1][x]!=v)
{
int p = fi(v);
addans(x,v,p,fa[1][p]);
}
}
}
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<n;i++)
{
int x,y;scanf("%d%d",&x,&y);
E[0][x].push_back(y);
E[0][y].push_back(x);
}
for(int i=1;i<n;i++)
{
int x,y;scanf("%d%d",&x,&y);
E[1][x].push_back(y);
E[1][y].push_back(x);
}
dfs(0,1,0);
dfs(1,1,0);
for(int i=1;i<=n;i++)
{
if(!check(0,i,fa[1][i]))
father[i]=i;
else
father[i]=fa[1][i];
}
solve(1,0);
cout<<tot<<endl;
for(int i=0;i<tot;i++,cout<<endl)
for(int j=0;j<4;j++)
printf("%d ",ans[i][j]);
}
Codeforces Round #345 (Div. 1) E. Clockwork Bomb 并查集的更多相关文章
- Codeforces Round #345 (Div. 2) E. Table Compression 并查集
E. Table Compression 题目连接: http://www.codeforces.com/contest/651/problem/E Description Little Petya ...
- Codeforces Round #345 (Div. 2) E. Table Compression 并查集+智商题
E. Table Compression time limit per test 4 seconds memory limit per test 256 megabytes input standar ...
- Codeforces Round #345 (Div. 1) C. Table Compression (并查集)
Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorith ...
- Codeforces Round #245 (Div. 2) B. Balls Game 并查集
B. Balls Game Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/430/problem ...
- Codeforces Round #603 (Div. 2) D. Secret Passwords 并查集
D. Secret Passwords One unknown hacker wants to get the admin's password of AtForces testing system, ...
- Codeforces Round #600 (Div. 2) D题【并查集+思维】
题意:给你n个点,m条边,然后让你使得这个这个图成为一个协和图,需要加几条边.协和图就是,如果两个点之间有一条边,那么左端点与这之间任意一个点之间都要有条边. 思路:通过并查集不断维护连通量的最大编号 ...
- Codeforces Round #600 (Div. 2) - D. Harmonious Graph(并查集)
题意:对于一张图,如果$a$与$b$连通,则对于任意的$c(a<c<b)$都有$a$与$c$连通,则称该图为和谐图,现在给你一张图,问你最少添加多少条边使图变为和谐图. 思路:将一个连通块 ...
- Codeforces Round #582 (Div. 3) G. Path Queries (并查集计数)
题意:给你带边权的树,有\(m\)次询问,每次询问有多少点对\((u,v)\)之间简单路径上的最大边权不超过\(q_i\). 题解:真的想不到用最小生成树来写啊.... 我们对边权排序,然后再对询问的 ...
- cf之路,1,Codeforces Round #345 (Div. 2)
cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅..... ...
随机推荐
- fork与vfork区别
1. 地址空间各段拷贝: fork: 内核为子进程生成新的地址空间结构,拷贝父进程的代码段,数据空间,堆,栈到自身的地址空间,但注意:子进程的代码段并不会分配物理空间,而是指向父进程的代码段物理空间, ...
- popup menu案例,无说明只代码
效果图: 布局文件, 展示列表的容器 <?xml version="1.0" encoding="utf-8"?> <LinearLayout ...
- [New learn] 网络基础-apache本地服务搭建(支持php)
1.简介 无网不利,无网不胜.对于移动应用来说离开网络那和咸鱼有什么分别?所以对于开发者来说更要学习好网络开发的技术. 2.搭建apache本地服务器 1.在finder中显示影藏的用户文件夹 fin ...
- [New learn]GCD的基本使用
https://github.com/xufeng79x/GCDDemo 1.简介 介绍GCD的使用,介绍多种队列与同步异步多种情况下的组合运行情况. 2.基本使用步骤 如果使用GCD则一般也就两个步 ...
- python manage.py 命令
在用命令django‐admin.py startproject <工程目录>建立一个django工程文件时,会生成一个manage.py文件,那么这个manage.py到底可以干嘛呢? ...
- java中的逻辑运算符,以及&与&&的区别,|与||的区别
原创,转载请留言联系 逻辑运算符: & 与 false&true=false:true&true=true:false&false=false 必须两个都是true才返 ...
- 自动化运维之saltstack的使用安装
SaltStack 简介 SaltStack是一个服务器基础架构集中化管理平台,具备配置管理.远程执行.监控等功能,基于Python语言实现,结合轻量级消息队列(ZeroMQ)与Python第三方模块 ...
- Bootstrap框架的简介
一.Bootstrap介绍 Bootstrap是Twitter开源的基于HTML.CSS.JavaScript的前端框架. 它是为实现快速开发Web应用程序而设计的一套前端工具包. 它支持响应式布局, ...
- 对angular.js的一点理解
最近一直在学习angular.js.不得不说和jquery相比有很大不同,有很多的不同点,之前也用过Knockout.js 但是两者还是有一定的区别的,首先knockout.js是基于Mvvm的,是几 ...
- javascript大神修炼记(4)——循环
读者朋友们大家好,今天,我们继续接着前面的内容讲,前们我们已经讲了条件分支,今天我们就讲循环,顾名思义就是,重复执行相同的操作,正常循环是受程序控制的,不正常的情况,就会出现死循环,那就是我们的代码中 ...