Socks CodeForces

Problem Description

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.

大致题意：你有n只袜子，要穿m天，每只袜子有一种颜色，你有k种颜料可以对袜子染色，接下来的m天每一天给你指定两只袜子穿，问你现在最少需要染色多少次才能使得接下来的m天每天穿的袜子颜色都相同。

思路： 我们可以将相关联的袜子用并查集找出一堆一堆的存放在vector中，然后我们需要将每一堆中的袜子颜色都染成相同的，那么每一堆最少需要的染色次数即该堆的袜子总数减去该堆相同颜色数量最多的袜子数，将每堆染色次数相加即是最终的答案。

代码如下

#include<iostream>#include<algorithm>#include<string>#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<cstring>#include<map>using namespace std;const int maxn=200005;int pre[maxn];int color[maxn];//记录每只袜子的颜色int jihe[maxn];//记录集合的编号int n,m,k;vector<int> v[maxn];int find(int x){   int r=x;   while (pre[r]!=r)   r=pre[r];   int i=x; int j;   while(i!=r)   {       j=pre[i];       pre[i]=r;       i=j;   }   return r;}void join(int x,int y){    int f1=find(x);    int f2=find(y);    if(f1!=f2)    {        pre[f2]=f1;    }}int main(){    int sum=0;    scanf("%d%d%d",&n,&m,&k);    for(int i=1;i<=n;i++)    pre[i]=i;    for(int i=1;i<=n;i++)    scanf("%d",&color[i]);    for(int i=1;i<=m;i++)    {        int x,y;        scanf("%d%d",&x,&y);        join(x,y);    }    int tol=1;    for(int i=1;i<=n;i++)    {        if(pre[i]==i)//给每个集合编号        {            jihe[i]=tol;            tol++;        }    }    for(int i=1;i<=n;i++)    {        v[jihe[find(i)]].push_back(color[i]);//记录每个集合的袜子的颜色    }    for(int i=1;i<tol;i++)    {        int Max=0;        map<int ,int> mp;        for(int j=0;j<v[i].size();j++)        {            mp[v[i][j]]++;//将v[i][j]颜色数量加一            if(mp[v[i][j]]>Max) Max=mp[v[i][j]];         }          sum+=v[i].size()-Max;    }    printf("%d\n",sum);   return 0;}