codeforces 691D D. Swaps in Permutation(dfs)

题目链接：

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, sopk = 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 apositions, 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

题意：

给一个数列,现在可以交换ai和bi,问能得到的最大的字典序的数列是什么;

思路：

把能交换位置的数放在同一个集合里;可以dfs找到,然后把这些数和位置排序对应就好了;不知道cf的数据怎么了;

AC代码：

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+10;
const int maxn=1005;
const double eps=1e-10;


int n,m,p[N],a[N],b[N],ans[N],cnt,vis[N];
vector<int>ve[N];
void dfs(int x)
{
    vis[x]=1;
    a[++cnt]=p[x];
    b[cnt]=x;
    int len =ve[x].size();
    For(i,0,len-1)
    {
        int y =ve[x][i];
        if(!vis[y]) dfs(y);
    }
}
int cmp(int x,int y)
{
    return x>y;
}
void solve()
{
    sort(a+1,a+cnt+1,cmp);
    sort(b+1,b+cnt+1);
    For(i,1,cnt)ans[b[i]]=a[i];
}

int main()
{

    read(n);read(m);
    For(i,1,n)read(p[i]);
    int u,v;
    For(i,1,m)
    {
        read(u);read(v);
        ve[u].push_back(v);
        ve[v].push_back(u);
    }
    For(i,1,n)
    {
        if(!vis[i])
        {
            cnt=0;
            dfs(i);
            solve(); 
        }
    }
    For(i,1,n)printf("%d ",ans[i]);

        return 0;
}