【codeforces 731C 】【并查集+贪心 或者dfs搞连通分支 】【有n只袜子,k种颜色,在m天中，左右脚分别穿下标为l,r的袜子，问最少修改几只袜子颜色，可以使每天穿的袜子左右两只都同色】

传送门：http://codeforces.com/contest/731/problem/C

描述：

C. Socks

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
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.

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 21 1 21 22 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.

题意：

有n只袜子（1~n），k种颜色（1~k），在m天中，左脚穿下标为l，右脚穿下标为r的袜子，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。

思路一：

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这堆袜子需要修改的颜色数，最后对每堆求和

代码一：

#include <bits/stdc++.h>
using  namespace  std;
#define rep(i,k,n) for(int i=k;i<=n;i++)

const int N=2e5+10;

int p
, col
, num
;
int tol, ans;
std::vector<int> v
;

int Find(int x){
return p[x] == x ? x : p[x] = Find(p[x]);
}

void unite(int x, int y){
x = Find(x);
y = Find(y);
if(x != y)
p[y] = x;
}

int  main(){
int n, m, k;
cin >> n >> m >> k;
rep(i, 1, n){
cin >> col[i];
}
rep(i, 1, n)p[i] = i;
int l, r;
rep(i, 1, m){
cin >> l >> r;
unite(l, r);//把颜色需要相同先放到一个集合中
}

rep(i, 1, n){
if(p[i] == i)num[i] = ++tol; // 给集合编号
}
rep(i, 1, n){
v[num[Find(i)]].push_back(col[i]); //每个集合中加入颜色
}
rep(i, 1, tol){
int sz = v[i].size();
int mx = 0;
std::map<int, int> mp;
rep(j, 0, sz - 1){
mp[v[i][j]] ++;
mx = max(mx, mp[v[i][j]]);//找到集合中颜色最多的 把剩下的元素变成该颜色即可
}
ans += (sz - mx);
}
cout << ans << endl;
return 0;
}

思路二：

每堆需要同色的袜子就是一个连通分支， 先建图，在每个连通分支中，把所有袜子的颜色都改成出现颜色次数最多的那个颜色，即该分支节点个数 - 最多出现次数

代码二：

#include <bits/stdc++.h>
using  namespace  std;
#define rep(i,k,n) for(int i=k;i<=n;i++)

const int N=3e5+10;

std::vector<int> g
;
int col
, vis
, cur, mx, ans;
std::map<int, int> num;

void dfs(int x){
if(vis[x])return ;
cur ++;
mx = max(mx, ++num[col[x]]);
vis[x] = 1;
for(auto y : g[x]){
dfs(y);
}
}

int  main(){
int n, m, k;
cin >> n >> m >> k;
rep(i, 1, n){
cin >> col[i];
}
int l, r;
rep(i, 1, m){
cin >> l >> r;
g[l].push_back(r);
g[r].push_back(l);
}
rep(i, 1, n){
if(! vis[i]){
cur = 0;
mx = 0;
dfs(i);
ans += (cur - mx);
num.clear(); //不要忘了
}
}
cout << ans << endl;
return 0;
}