#452. 序列操作

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
const int N = 1e6 + 10;
struct Node {
    int op;
    int x, y;
}t[N];
int n;
int a[N], q;
bool vis[N];
void solve() {
    cin >> n >> q;
    for (int i = 1;  i<= n ;i ++)
        a[i] = read();
    for (int i = 1; i<= q; i ++) {
        t[i].op = read();
        if (t[i].op == 1) {
            t[i].x = read();
            t[i].y = read();
        }
        else {
            t[i].x = read();
        }
    }
    int mx = 0;
    for (int i = q; i >= 1;  i--) {
        if (t[i].op == 2) {
            mx = max (t[i].x, mx);
        }
        else if (!vis[t[i].x] && t[i].op == 1){
            a[t[i].x] = max (mx, t[i].y);
            
            vis[t[i].x] = 1;
        }
    }
    for (int i = 1; i <= n; i ++)
        if (!vis[i]) a[i] = max(mx, a[i]);
    for (int i = 1; i <= n; i ++) {
        cout << a[i] << ' ';
    }
    
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

关键在于处理第二种操作。 我们可以倒着求。这样如果当前操作是1，记录最大值和当前要改成的数字进行比较，取最大值，然后标记为已经修改，并且不能够再被修改。因为我们是倒着求的，每次的结果都是最新的结果。