Codeforces Round #576 (Div. 2) D. Welfare State||Codeforces Round #576 (Div. 1) B Welfare State

 Welfare State

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a country with n citizens. The i-th of them initially has ai money. The government strictly controls the wealth of its citizens. Whenever a citizen makes a purchase or earns some money, they must send a receipt to the social services mentioning the amount of money they currently have.

Sometimes the government makes payouts to the poor: all citizens who have strictly less money than x are paid accordingly so that after the payout they have exactly x money. In this case the citizens don't send a receipt.

You know the initial wealth of every citizen and the log of all events: receipts and payouts. Restore the amount of money each citizen has after all events.

Input

The first line contains a single integer n (1≤n≤2e5) — the numer of citizens.

The next line contains n integers a1, a2, ..., an (0≤ai≤1e9) — the initial balances of citizens.

The next line contains a single integer q(1≤q≤2e5) — the number of events.

Each of the next q lines contains a single event. The events are given in chronological order.

Each event is described as either 1 p x (1≤p≤n, 0≤x≤1e9), or 2 x (0≤x≤1e9). In the first case we have a receipt that the balance of the p-th person becomes equal to x. In the second case we have a payoff with parameter x.

Output

Print n integers — the balances of all citizens after all events.

题意：

有个国家想劫富济贫，所以会给你n个人的财富ai，然后有两种操作，1，x，y的意思是把ax人的资产修改成y；2，x的意思是将所有资产低于x的人的资产全部改成x。

输出最后所有人的资产

思路：

我们可以分别去想每一个公民，一个公民的最终资产就是，max（最后一次的1操作的值，之后的最大的2操作的值），所以输入的每一个1，修改资产并用pos数组记录这个人这次修改的操作序号，每次2操作，用一个sec数组记录这次的值，然后从序号q向1把最大值更新给前面，因为只有之后的最大值才有用。然后1...n遍历max(a[i],sec[pos[i]+1]

 

代码：

#include<iostream>
using namespace std;
const int N=2e5+10;
int n,a[N],q,o,p,x,pos[N],sec[N];
int main() {
    cin>>n;
    for(int i=1; i<=n; i++)cin>>a[i];
    cin>>q;
    for(int i=1; i<=q; i++) {
        cin>>o;
        if(o==1) {
            cin>>p>>x;
            a[p]=x;//修改下标为p的人的资产 
            pos[p]=i;//记录下标为p的人资产修改的最后一次操作序号i 
        } else cin>>sec[i];//记录操作2的值 
    }
    for(int i=q; i>=1; i--)
        sec[i] = max(sec[i],sec[i+1]);//把操作2的最大值向回传递 
    for(int i=1; i<=n; i++)
        cout<<max(a[i],sec[pos[i]+1])<< " ";//max（最后一次的1操作的值，之后的最大的2操作的值） 
    return 0;
}