题干：
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.
Input
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.
Output
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.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
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.
题目大意：
有n只袜子，m天，k个颜色。（n,m,k<=2e5）
第一行n个数，代表每只袜子的颜色。接下来m行给出m天中，每天穿的两只袜子的编号，每只袜子可能不同颜色，现在要让每天穿的两只袜子是相同颜色的，要重新染色的袜子数最少是多少。
解题报告：
直接并查集，然后找出并查集中的颜色最多的袜子是多少，因为并查集后属于不同集合的元素一定没有在m天中同时出现过，所以集合之间互不影响，所以这样做，也不会出现一只袜子被重复染色多次的情况。
因为这里只要最大值，所以不需要对每个集合都先排序然后找最大，而是直接开数组维护就好了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n,m,k;
int c[MAX],f[MAX],cnt[MAX];
int getf(int v) {return f[v] == v ? v : f[v] = getf(f[v]);
}
void merge(int u,int v) {int t1 = getf(u),t2 = getf(v);f[t2] = t1;
}
vector<int> vv[MAX];
int mx,ans;
int main()
{cin>>n>>m>>k;for(int i = 1; i<=n; i++) scanf("%d",c+i),f[i]=i;for(int u,v,i = 1; i<=m; i++) {scanf("%d%d",&u,&v);merge(u,v);}for(int i = 1; i<=n; i++) getf(i);for(int i = 1; i<=n; i++) {vv[f[i]].pb(c[i]);}for(int i = 1; i<=n; i++) {if(vv[i].size() <= 1) continue;mx = 0;for(auto x : vv[i]) cnt[x]++,mx = max(mx,cnt[x]); for(auto x : vv[i]) cnt[x] = 0;//initans += vv[i].size() - mx;}cout << ans << endl;return 0 ;
}
