Codeforces Round #376 (Div. 2) C. Socks

可能英语水平比较差，这个题我刚读起来是懵逼的。

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 kcolors.

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 31 2 31 22 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.

题目大意：

题意：你有n双袜子，要穿m天，每个袜子都有一种颜色，你有k种颜料，每天给你指定两双袜子，你每天要穿这两双袜子出门，但是这两双袜子必须颜色一样，你可以对任意一个袜子进行染色，这样他们就一样了，问你：穿下这m天，最少要涂多少种颜色。

思路：你m天穿的袜子，如果有某几天穿的袜子是一样的，那么这几天的袜子都必须一样。这就是并查集了。。。用并查集把那一部分所有的袜子都变成一种颜色，最少的涂法就是用这个集合所有的袜子减去颜色最多的袜子就是最少的涂法，这样的集合可能在这m天有好几个，就分别统计每个集合的涂法。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<algorithm>using namespace std;int a[200005],col[200005],b[200005];vector<int> ve[200005];int find(int num){    if(num==a[num])        return num;    else return a[num]=find(a[num]);}void insert(int l,int r){    a[find(l)]=find(r);}int main(){    int n,m,k;    int i,j;    int ans;    set<int>st;    scanf("%d %d %d",&n,&m,&k);    int len;    map<int,int>mp;    int maxn;    int l,r;    for(i=1;i<=n;i++)        a[i]=i;    for(i=1;i<=n;i++){        scanf("%d",&col[i]);    }    for(i=1;i<=m;i++){        scanf("%d %d",&l,&r);        insert(l,r);    }    len=0;    for(i=1;i<=n;i++){        if(find(i)==i){            b[i]=len;            len++;        }    }    for(i=1;i<=n;i++){        ve[b[find(i)]].push_back(col[i]);    }    ans=0;    for(i=0;i<len;i++){        maxn=0;        for(j=0;j<ve[i].size();j++){            mp[ve[i][j]]++;            if(maxn<mp[ve[i][j]])                maxn=mp[ve[i][j]];        }        ans+=ve[i].size()-maxn;        mp.clear();    }    printf("%d\n",ans);    return 0;}