time limit per test5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a permutation of the numbers 1, 2, …, n and m pairs of positions (aj, bj).

At each step you can choose a pair from the given positions and swap the numbers in that positions. What is the lexicographically maximal permutation one can get?

Let p and q be two permutations of the numbers 1, 2, …, n. p is lexicographically smaller than the q if a number 1 ≤ i ≤ n exists, so pk = qk for 1 ≤ k < i and pi < qi.

Input
The first line contains two integers n and m (1 ≤ n, m ≤ 106) — the length of the permutation p and the number of pairs of positions.

The second line contains n distinct integers pi (1 ≤ pi ≤ n) — the elements of the permutation p.

Each of the last m lines contains two integers (aj, bj) (1 ≤ aj, bj ≤ n) — the pairs of positions to swap. Note that you are given a positions, not the values to swap.

Output
Print the only line with n distinct integers p’i (1 ≤ p’i ≤ n) — the lexicographically maximal permutation one can get.

Example
input
9 6
1 2 3 4 5 6 7 8 9
1 4
4 7
2 5
5 8
3 6
6 9
output
7 8 9 4 5 6 1 2 3

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
//#include<vector>
#include<functional>
//#include<Windows.h>
using namespace std;
const int maxn = 1e6 + 50;
//#define INF 0x3f3f3f3f
#define ll long long int
priority_queue<int> Q[maxn];
int f[maxn];
int a[maxn];
int FIND(int x)
{
    if (x == f[x])
        return f[x];
    else
        return f[x] = FIND(f[x]);
}
void getf(int x, int y)
{
    int fx = FIND(x);
    int fy = FIND(y);
    if (fx != fy)
    {
        f[fx] = fy;
    }
}
int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    //memset(a, 0, sizeof(a));
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        f[i] = i;
    }
    /*for (int i = 1; i <= n; i++)
    {

    }*/
    while (m--)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        getf(x, y);
    }
    for (int i = 1; i <= n; i++)
    {
        Q[f[FIND(i)]].push(a[i]);//WA點
    }
    for (int i = 1; i <= n; i++)
    {
        if (i != 1)
            printf(" ");
        printf("%d", Q[f[i]].top());
        Q[f[i]].pop();
    }
    printf("\n");
    //system("pause");
    return 0;
}