D. Welfare State
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.
The first line contains a single integer n (1≤≤2⋅1051≤n≤2⋅105) — the numer of citizens.
The next line contains n integers 1a1, 2a2, ..., an (0≤≤1090≤ai≤109) — the initial balances of citizens.
The next line contains a single integer q (1≤≤2⋅1051≤q≤2⋅105) — 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≤≤1≤p≤n, 0≤≤1090≤x≤109), or 2 x (0≤≤1090≤x≤109). 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.
Print n integers — the balances of all citizens after all events.
4
1 2 3 4
3
2 3
1 2 2
2 1
3 2 3 4
5
3 50 2 1 10
3
1 2 0
2 8
1 3 20
8 8 20 8 10
In the first example the balances change as follows: 1 2 3 4 →→ 3 3 3 4 →→ 3 2 3 4 →→ 3 2 3 4
In the second example the balances change as follows: 3 50 2 1 10 →→ 3 0 2 1 10 →→ 8 8 8 8 10 →→ 8 8 20 8 10
哭了哭了~~
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=200010;
int a[maxn],last[maxn],b[maxn];//a为原数组,last[i]记录修改i节点的最后一个位置,b[i]代表从i时间点往后最大的补充x值
int main()
{
int n,q;
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i];
cin>>q;
for(int i=1;i<=q;i++){
int op,x,y;
cin>>op;
if(op==1){
cin>>x>>y;
a[x]=y;//修改原数组值
last[x]=i;//记录修改x下标的最后一个位置i
}
else{
cin>>b[i];//输入i时间点的补充值b[i]
}
}
for(int i=q-1;i>=0;i--)b[i]=max(b[i],b[i+1]);
for(int i=1;i<=n;i++)cout<<max(a[i],b[last[i]])<<" ";
cout<<endl;
return 0;
}
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