C. Socks
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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.

題意：給出N，M，K代表有N只襪子，要穿M天，最多有K種顏色，接著一行N個整數ci，代表第i只襪子顏色為ci，接著M行，每行給出兩個整數li，ri代表第i天穿的襪子編號。為了保證每天穿的兩雙襪子顏色相同，需要對某些襪子進行染色，求出需要染色襪子數量的最小值。

思路：根據兩隻襪子是否在同一天穿，可以劃定數個集合，對於每個集合除了出現次數最多的那個顏色，其他顏色的襪子都需要染成出現次數最多的顏色。

剛開始用聯通圖，深搜遍歷每一塊區域，後來發現這不就是並查集。。。。。

程式碼

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<map>
using namespace std;
const int maxn=200005;
int father[maxn*10];//第i號襪子是和father[i]在同一集合的
int color[maxn*10];//第i號襪子的顏色是color[i]
//int num[maxn];//第i種顏色的襪子在當前集合中出現了num[i]次
int Find_father(int x)
{
    if(x!=father[x])
        father[x]=Find_father(father[x]);
    return father[x];
}
void Merge(int x,int y)
{
    int flag_x=Find_father(x);
    int flag_y=Find_father(y);
    if(flag_x!=flag_y)
        father[flag_x]=flag_y;
}
vector<int>group[maxn];//集合i擁有的顏色
int main()
{
    int N,M,K;
    scanf("%d%d%d",&N,&M,&K);
    for(int i=1; i<=N; i++)
    {
        scanf("%d",&color[i]);
        father[i]=i;
    }
    while(M--)
    {
        int left,right;
        scanf("%d%d",&left,&right);
        Merge(left,right);
    }

    for(int i=1; i<=N; i++)
    {
        int flag_i=Find_father(i);
        group[flag_i].push_back(color[i]);
    }
    int result=0;
    for(int i=1; i<=N; i++)
    {
        int len=(int)group[i].size();
        if(len!=0)
        {
            map<int,int>mp;
//            memset(num,0,sizeof(num));
            int max_num=0;
            for(int j=0; j<len; j++)
            {
                mp[group[i][j]]++;
                max_num=max(max_num,mp[group[i][j]]);
            }
            max_num=len-max_num;
            result+=max_num;
        }
    }
    printf("%d\n",result);
    return 0;
}