题目链接

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;

}