CodeForces731-C.Socks-并查集
2 seconds
256 megabytes
standard input
standard output
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
3 2 3
1 2 3
1 2
2 3
2
3 2 2
1 1 2
1 2
2 1
0
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
迷之题面。。。
大体题意就是一个人穿袜子,左脚的袜子,右脚的袜子,然后要穿颜色相同的袜子,不同的怎么办呢,染色,问最少染颜色的次数。
数据第一行三个数代表有几个袜子,穿几天,几种颜色,第二行是袜子颜色,下面几行就是这些天里每天走右脚穿的袜子。例如第一组数据,1 2就是第一天左脚1右脚2,2 3是第二天左脚2,右脚3,因为两天里都有2,所以第一天染1,第二天染3,所以为2。
并查集,找出集合中穿的颜色最多的,把剩下的都染色。
代码:
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+;
int hh[N],f[N];
vector<int>p[N]; //类似一个动态数组p[N]
int Find(int x){ //查找根节点
if(f[x]==x)
return f[x];
else
return f[x]=Find(f[x]);
}
void gg(int a,int b){
int x=Find(a);
int y=Find(b);
if(x!=y) //压缩
f[x]=y;
}
int main(){
int n,m,k,l,r,ans=;
scanf("%d%d%d",&n,&m,&k);
for(int i=;i<=n;i++){
scanf("%d",&hh[i]); //袜子颜色
f[i]=i;
}
for(int i=;i<=m;i++){
scanf("%d%d",&l,&r); //左右脚袜子
gg(l,r);
}
for(int i=;i<=n;i++)
p[Find(i)].push_back(hh[i]); //将颜色这个元素添加进去
for(int i=;i<=n;i++){
if(p[i].size()<=)continue;
int maxx=-;
map<int,int> mp;
mp.clear(); //清空
for(int j=;j<p[i].size();j++){
mp[p[i][j]]++;
maxx=max(maxx,mp[p[i][j]]);
}
ans+=p[i].size()-maxx;
}
printf("%d\n",ans);
return ;
}
(T▽T)
CodeForces731-C.Socks-并查集的更多相关文章
- CodeForces 731C C - Socks 并查集
Description Arseniy is already grown-up and independent. His mother decided to leave him alone for m ...
- Codeforces 731C Socks 并查集
题目:http://codeforces.com/contest/731/problem/C 思路:并查集处理出哪几堆袜子是同一颜色的,对于每堆袜子求出出现最多颜色的次数,用这堆袜子的数目减去该值即为 ...
- CF731C Socks并查集(森林),连边,贪心,森林遍历方式,动态开点释放内存
http://codeforces.com/problemset/problem/731/C 这个题的题意是..小明的妈妈给小明留下了n只袜子,给你一个大小为n的颜色序列c 代表第i只袜子的颜色,小明 ...
- Codeforces Round #376 (Div. 2) C. Socks —— 并查集 + 贪心
题目链接:http://codeforces.com/contest/731/problem/C 题解: 1.看题目时,大概知道,不同的袜子会因为要在同一天穿而差生了关联(或者叫相互制约), 其中一条 ...
- Codeforces 731C:Socks(并查集)
http://codeforces.com/problemset/problem/731/C 题意:有n只袜子,m天,k个颜色,每个袜子有一个颜色,再给出m天,每天有两只袜子,每只袜子可能不同颜色,问 ...
- 并查集【CF731C】Socks
Description 有\(n\)只袜子,\(k\)种颜色,在\(m\)天中,问最少修改几只袜子的颜色,可以使每天要穿的袜子颜色相同. Input 第一行\(n,m,k\)分别对应题目描述. 接下来 ...
- CodeForces 731C Socks (DFS或并查集)
题意:有n只袜子,k种颜色,在m天中,问最少修改几只袜子的颜色,可以使每天穿的袜子左右两只都同颜色. 析:很明显,每个连通块都必须是同一种颜色,然后再统计最多颜色的就好了,即可以用并查集也可以用DFS ...
- Codeforces 731 C.Socks-并查集+STL(vector+map)
C. Socks time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...
- codeforces_731C_[dfs][并查集]
C. Socks time limit per test 2 seconds memory limit per test 256 megabytes input standard input outp ...
随机推荐
- 简单介绍什么是协程及其在ES6中的实现方式
协程,英文名coroutine,是一种执行过程可以被暂停和恢复的方法.各个协程之间相互协作完成一个任务. 让我们来看一个关于发挥协程作用的例子.假定我们有一个生产者和消费者的关系,生产者创建物品并将物 ...
- 【Zookeeper】源码分析目录
Zookeeper源码分析目录如下 1. [Zookeeper]源码分析之序列化 2. [Zookeeper]源码分析之持久化(一)之FileTxnLog 3. [Zookeeper]源码分析之持久化 ...
- ArcGIS 网络分析[8.3] 设置IDENetworkDataset的属性及INetworkDataset的对比/创建网络数据集
创建网络数据集就得有各种数据和参数,这篇文章很长,慎入. 网络分析依赖于网络数据集的质量,这句话就在这里得到了验证:复杂.精确定义. 本节目录如下: 1. INetworkDataset与IDENet ...
- 安装MySQL时提示3306端口已被占用的解决方案
之前安装过mysql,用的好好的,但是今天开启服务时报异常,无法启动.为省事,于是想到卸载重装,安装过程中发现3306已经被占用,这也是一开始服务无法启动的原因. 看到有人说用fport查看端口号,于 ...
- git 分支操作
查看git分支: git fetch刷新git git branch -a 列出所有的分支 git checkout origin/要切换的分支 git branch -r 查看远程分支 git c ...
- ADO.NET访问数据库
1:ADO.NET数据库的方法和技术 2:ADO.NET的主要组成: 1>DataSet(数据集)-----独立于数据间的数据访问 2>.NETFramework(数据提供程序)----- ...
- Ruby学习之对象模型
这两周工作内容较多,平时自己也有点不在状态,学的东西有点少了,趁着现在还有点状态,赶紧复习一下之前学习的Ruby吧. Ruby是我真正开始接触动态语言魅力的第一个语言,之前虽然也用过且一直用perl. ...
- Validation of viewstate MAC failed 解决办法
大部分人都说是在页里或web.config里加EnableEventValidation="false" EnableViewStateMac="false" ...
- MySQL 优化实施方案
1.1 前言 在进行MySQL的优化之前必须要了解的就是MySQL的查询过程,很多的查询优化工作实际上就是遵循一些原则让MySQL的优化器能够按照预想的合理方式运行而已.更多关于MySQL查询相关参照 ...
- Elasticsearch强制重置未分配的分片(unassigned)
强制重置未分片的分片,这个问题源自于Elasticsearch维护中,Node意外退出的场景. 意外退出后Elasticsearch由于网络原因或者jvm性能压力,未能短时间内分配分片. 看一下分片的 ...