codeforces 691D（数据结构）

D. Swaps in Permutation

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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

思路：可以先将所有可以交换的区间用并查集合并，然后用多个大根堆维护并查集的根可以的数字，最后只需依次将根可以的最小数字放到数组中即可，放一个数字则删除根元素。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
using namespace std;
priority_queue<int> ans[1000001];
int n,m,p[1000001],father[1000001];
int ask(int x)
{
    if (father[x]==x) return x;
    father[x]=ask(father[x]);
    return father[x];
}
void hebin(int x,int y)
{
    int p=ask(x);
    int q=ask(y);
    if (p!=q) father[p]=q;
}
int main()
{
    scanf("%d%d",&n,&m);
    int i;
    for (i=1;i<=n;i++)
    {
        scanf("%d",&p[i]);
        father[i]=i;
    }
    while (m--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        hebin(x,y);
    }
    for (i=1;i<=n;i++) ans[father[ask(i)]].push(p[i]);
    for (i=1;i<=n;i++)
    {
        printf("%d ",ans[father[i]].top());
        ans[father[i]].pop();
    }
    return 0;
}