이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//JOIOC 2021 p2 finanicial
#include <bits/stdc++.h>
#define le node+node
#define ri node+node+1
#define mid (l+r)/2
#define pi pair<int,int>
using namespace std;
const int inf = 3e5+9;
int n,d,ans,cnt,a[inf],dp[inf];
map<int,int> mp;
int tree[inf<<2];
void update(int node,int l,int r,int idx,int val,int del){
if(l == r){
if(del)
return void(tree[node] = val);
return void(tree[node] = max(tree[node],val));
}
if(idx<=mid)
update(le,l,mid,idx,val,del);
else
update(ri,mid+1,r,idx,val,del);
tree[node] = max(tree[le],tree[ri]);
}
int query(int node,int l,int r,int x,int y){
if(l>r || r<x || l>y || x>y)
return 0;
if(l>=x && r<=y)
return tree[node];
return max(query(le,l,mid,x,y),query(ri,mid+1,r,x,y));
}
deque<pair<int,int> > dq;//pair<time,val>
set<int> vals;
int main(){
scanf("%d%d",&n,&d);
for(int i=1;i<=n;i++)
scanf("%d",a+i),mp[a[i]];
for(auto &o:mp)
o.second = ++cnt;
for(int i=1;i<=n;i++)
a[i] = mp[a[i]];
for(int i=1;i<=n;i++){
dp[i] = query(1,1,cnt,1,a[i]-1)+1;
update(1,1,cnt,a[i],dp[i],0);
vals.insert(a[i]);
if(i == n)
ans = tree[1];
//cout<<dp[i]<<" ";
while(!dq.empty() && dq.back().second >= a[i])
dq.pop_back();
dq.push_back(pi(i,a[i]));
while(dq.front().first <= i-d)
dq.pop_front();
int LimitToDelete = dq.front().second-1;
while(!vals.empty() && *vals.begin() <= LimitToDelete){
update(1,1,cnt,*vals.begin(),0,1);
vals.erase(vals.begin());
}
}
printf("%d\n",ans);
}
컴파일 시 표준 에러 (stderr) 메시지
Main.cpp: In function 'int main()':
Main.cpp:37:10: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
37 | scanf("%d%d",&n,&d);
| ~~~~~^~~~~~~~~~~~~~
Main.cpp:39:14: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
39 | scanf("%d",a+i),mp[a[i]];
| ~~~~~^~~~~~~~~~
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