Codeforces Round #376 (Div. 2) C. Socks 并查集+贪心、图论

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 ofk 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 mdays.

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.

Source

Codeforces Round #376 (Div. 2)

My Solution

 并查集+贪心、图论

在读入的时候直接把有边相连的点维护到一个集合里，最后对于处理出的森林，可以用map<int, map<int, int> > mp;维护，即mp[i][j]表示以 i 为根的树上颜色 j 出现的次数，

然后对于每颗树，找出树上出现的次数最多的结点颜色相同的颜色， ans += (该树的总结点数) - 出现的次数最多的结点颜色相同的颜色的结点个数

复杂度 O(n)

#include <iostream>
#include <cstdio>
#include <map>
#include <cstring>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + 8;

int father[maxn], _rank[maxn];
bool flag;
inline void DisjointSet(int n)
{
for(int i = 0; i <= n; i++){
father[i] = i;
}
}

int _find(int v)
{
return father[v] = father[v] == v ? v : _find(father[v]);
}

void _merge(int x, int y)
{
int a = _find(x), b = _find(y); //
if(_rank[a] < _rank[b]){
father[a] = b;
}
else{
father[b] = a;
if(_rank[a] == _rank[b]){
_rank[a]++;
}
}
}

int val[maxn];
map<int, map<int, int> > mp;
map<int, int> ok;

int main()
{
#ifdef LOCAL
freopen("c.txt", "r", stdin);
//freopen("c.out", "w", stdout);
int T = 4;
while(T--){
#endif // LOCAL
ios::sync_with_stdio(false); cin.tie(0);

int n, m, k, u, v;
cin >> n >> m >> k;
DisjointSet(n);
for(int i = 1; i <= n; i++){
cin >> val[i];
}
for(int i = 0; i < m; i++){
cin >> u >> v;
ok[u]++;
ok[v]++;
if(_find(u) != _find(v)){
//cout << "?";
_merge(_find(u), _find(v));
}
}

for(auto i = ok.begin(); i != ok.end(); i++){
mp[_find(i->first)][val[i->first]]++;
}

int ans = 0, sum = 0, maxv = 0;
for(auto i = mp.begin(); i != mp.end(); i++){
sum = 0; maxv = 0;
for(auto j = (i->second).begin(); j != (i->second).end(); j++){
//cout << j->second << " ";
sum += j->second;
maxv = max(maxv, j->second);
}
//cout << endl;
ans += (sum - maxv);
}

cout << ans << endl;

#ifdef LOCAL
mp.clear();
cout << endl;
}
#endif // LOCAL
return 0;
}

  Thank you!

                                                                                                           
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