Educational Codeforces Round 14 D. Swaps in Permutation

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            secondconst 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            secondconst 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;}