CodeForces731C 并查集

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.

这题是见过的最奇怪的并查集….
思路是把小集合并入大集合中….
这么做的原因是即便是同颜色也不一定同集合..
很容易让人搞混
还有一个关键点是
如果一个点被用了两次
那么他必然最后的结果是第二种颜色..
这样的时候就会发现，他的第一种状态在染成第二种颜色的时候没法保证
所以将他爹也染成这个颜色，由于在同一个集合内，有一个没法保证其他的就都没法保证….
所以全体染成这个色
这样就简化成
如何合并集合操作最少了…

#include<iostream>
#include<cstdio>
#include<memory.h>
#include<vector>
#include<algorithm>
using namespace std;
vector <int> q[200001];
int tu[250000], die[250000], jilu[250000];
int zhao(int x)
{
    if (x == die[x])return x;
    else return die[x] = zhao(die[x]);
}
int main()
{
    int n, m, k;
    cin >> n >> m >> k;
    for (int a = 1;a <= n;a++)scanf("%d", &tu[a]), die[a] = a;
    int sum = 0;
    int e, w;
    for (int a = 1;a <= m;a++)
    {
        scanf("%d%d", &e, &w);
        int qd = zhao(e), wd = zhao(w);
        die[qd] = wd;
    }
    for (int a = 1;a <= n;a++)q[zhao(die[a])].push_back(a);
    for (int a = 1;a <= n;a++)
    {
        if (!q[a].size())continue;
        int gagaga = 0;
        for (int b = 0;b<q[a].size();b++)
        {
            jilu[tu[q[a][b]]]++;
            gagaga = max(gagaga, jilu[tu[q[a][b]]]);
        }
        for (int b = 0;b<q[a].size();b++)
        {
            jilu[tu[q[a][b]]]--;
        }
        sum += q[a].size();
        sum -= gagaga;
    }
    cout << sum << endl;
    return 0;
}