题目链接

题意：




题解：














AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
#define mem(a, b) memset(a, b, sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod = 1e9 + 7;
const int mod = 998244353;

const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 3e3 + 5;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
int a[maxn], mn[maxn], col[maxn];
vector<int> g[maxn];
bool dfs(int u, int c) {
    col[u] = c;
    for (auto v : g[u]) {
        if (col[v] == c) return false;
        if (col[v] == -1 && !dfs(v, c ^ 1)) return false;
    }
    return true;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    mn[n + 1] = inf;
    for (int i = n; i; i--)
        mn[i] = min(mn[i + 1], a[i]);
    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
            if (a[i] < a[j] && mn[j + 1] < a[i]) {
                g[i].pb(j);
                g[j].pb(i);
            }
    mem(col, -1);
    bool ok = 1;
    for (int i = 1; i <= n; i++)
    if (col[i] == -1 && !dfs(i, 0)) {
        ok = 0;
        break;
    }
    if (!ok) {cout << 0; return 0;}
    stack<int> s1, s2;
    int now = 1;
    for (int i = 1; i <= n; i++) {
        if (!col[i]) s1.push(a[i]), cout << "a ";
        else s2.push(a[i]), cout << "c ";
        while (1) {
            if (!s1.empty() && s1.top() == now) s1.pop(), cout << "b ";
            else if (!s2.empty() && s2.top() == now) s2.pop(), cout << "d ";
            else break;
            now++;
        }
    }
    return 0;
}