分析：一些边把各个节点连接成了一颗颗树。因为每棵树上的边可以走任意次，所以不难想出要字典序最大，就是每棵树中数字大的放在树中节点编号比较小的位置。
我用了极为暴力的方法，先dfs每棵树，再用了优先队列。我估计最大复杂度约在O(Nlog(N))，理论上应该跑不过。因为再cf上做题，看见5s时限，强行上了。很侥幸，在4秒的时候过了= =。
/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define offcin ios::sync_with_stdio(false)
#define sigma_size 26
#define lson l,m,v<<1
#define rson m+1,r,v<<1|1
#define slch v<<1
#define srch v<<1|1
#define sgetmid int m = (l+r)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define pb push_back
#define fi first
#define se second
const int INF = 0x3f3f3f3f;
const LL INFF = 1e18;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const int maxmat = 10;
const ull BASE = 31;
/*****************************************************/
const int maxn = 1e6 + 5;
std::vector<int> G[maxn];
int N, M;
int p[maxn];
bool vis[maxn];
struct Node {
int val;
bool operator <(const Node &rhs) const {
return val > rhs.val;
}
};
priority_queue<Node> q;
priority_queue<int> ans;
void dfs(int u, int fa) {
if (vis[u]) return;
vis[u] = true;
ans.push(p[u]);
q.push((Node){u});
for (int i = 0; i < G[u].size(); i ++) {
int v = G[u][i];
if (v == fa) continue;
dfs(v, u);
}
}
int main(int argc, char const *argv[]) {
cin>>N>>M;
mem(vis, false);
for (int i = 1; i <= N; i ++) scanf("%d", p + i);
for (int i = 0; i < M; i ++) {
int u, v;
scanf("%d%d", &u, &v);
G[u].pb(v);
G[v].pb(u);
}
for (int i = 1; i <= N; i ++) {
if (!vis[i]) dfs(i, -1);
while (!ans.empty()) {
int tmp = ans.top(); ans.pop();
Node x = q.top(); q.pop();
int id = x.val;
p[id] = tmp;
}
}
for (int i = 1; i <= N; i ++) cout<<p[i]<<" ";
return 0;
}正解没有那么暴力啊。正解是通过并查集来区分树。因为并查集的原因，遍历点的时候一定是从小到大的，省去了对位置的排序，复杂度又降了一个常数。这里只需要对各个树中的值排序就行了。复杂度比我的方法小了太多。用vector就可以了。
/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define offcin ios::sync_with_stdio(false)
#define sigma_size 26
#define lson l,m,v<<1
#define rson m+1,r,v<<1|1
#define slch v<<1
#define srch v<<1|1
#define sgetmid int m = (l+r)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define pb push_back
#define fi first
#define se second
const int INF = 0x3f3f3f3f;
const LL INFF = 1e18;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const int maxmat = 10;
const ull BASE = 31;
/*****************************************************/
const int maxn = 1e6 + 5;
int par[maxn], p[maxn];
std::vector<int> pos[maxn];
std::vector<int> num[maxn];
int N, M;
void init() {
for (int i = 1; i <= N; i ++) par[i] = i;
}
int findpar(int x) {
return par[x] = (par[x] == x ? x : findpar(par[x]));
}
void unite(int x, int y) {
x = findpar(x), y = findpar(y);
if (x == y) return;
par[x] = y;
}
int main(int argc, char const *argv[]) {
cin>>N>>M;
init();
for (int i = 1; i <= N; i ++) scanf("%d", p + i);
for (int i = 0; i < M; i ++) {
int u, v;
scanf("%d%d", &u, &v);
unite(u, v);
}
for (int i = 1; i <= N; i ++) {
int x = findpar(i);
pos[x].pb(i);
num[x].pb(-p[i]);
}
for (int i = 1; i <= N; i ++) {
sort(num[i].begin(), num[i].end());
for (int j = 0; j < pos[i].size(); j ++) {
int id = pos[i][j];
p[id] = -num[i][j];
}
}
for (int i = 1; i <= N; i ++) cout<<p[i]<<" ";
return 0;
}