Submission #940084

#	Time	Username	Problem	Language	Result	Execution time	Memory
940084		vjudge1	Global Warming (CEOI18_glo)	C++17	100 / 100	611 ms	171768 KiB

glo

#include <bits/stdc++.h>
 
#define ff first
#define ss second
#define pb push_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define int long long
#define rnd(l, r) uniform_int_distribution<int>(l, r)(rng)
 
using namespace std;

void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);} 
int pow(int a,int b,int m){int ans=1;while(b){if(b&1){ans=(ans*a)%m;}b>>=1;a=(a*a)%m;}return ans;}
int binpow(int a,int b){int ans=1;while(b){if(b&1){ans=(ans*a);}b>>=1;a=(a*a);}return ans;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int inf = 2e9 + 10, N = 2e5 + 100;
 
struct node{
    int mn;
    node *l, *r;
    node(){
        mn = 0;
        l = nullptr;
        r = nullptr;
    }
};
void update(node *&p, int pos, int val, int tl = -inf, int tr = inf){
    if(p == nullptr){
        p = new node();
    }
    if(tl == tr){
        p -> mn = val;
        return;
    }
    int mid = (tl + tr) >> 1;
    if(pos > mid)
        update(p -> r, pos, val, mid + 1, tr);
    else
        update(p -> l, pos, val, tl, mid);
    int left = 0;
    if(p -> l != nullptr)
        left = p -> l -> mn;
    int right = 0;
    if(p -> r != nullptr)
        right = p -> r -> mn;
    p -> mn = max(left, right);
}

int get(node *p, int l, int r, int tl = -inf, int tr = inf){
	if(p == nullptr)return 0;
    if(tl > r || tr < l)
        return 0;
    if(tl >= l && tr <= r)
        return p -> mn;
    int mid = (tl + tr) >> 1;
    return max(get(p -> r, l, r, mid + 1, tr), get(p -> l, l, r, tl, mid));
}
int a[N];
main(){
	iostream::sync_with_stdio(false);	
    cin.tie(nullptr);
    cout.tie(nullptr);
	int n, d;
	cin >> n >> d;
	node *root = new node(), *root2 = new node();
	int ans = 0;
	for(int i = 1 ; i<= n; i++){
		cin >> a[i];
		int j = get(root, 0, a[i] + d - 1);
		int j2 = get(root2, 0, a[i] - 1);
		int x2 = max({get(root2, a[i], a[i]), j + 1, j2 + 1});
		update(root2, a[i], x2);
		//cout << i <<" "<< x2 << endl;
		ans = max(ans, x2);
		
		int g = get(root, 0, a[i] - 1);
		ans = max(g + 1, ans);
		int x = get(root, a[i], a[i]);
		if(x < g + 1)
			update(root, a[i], g + 1);
		
	}
	cout << ans << endl;
}	
/*
 * Before implementing the problem:
	
	Do I understand the problem correctly?
	Which places are tricky? What do I need to remember to implement them correctly?
	Which place is the heaviest by implementation? Can I do it simpler?
	Which place is the slowest? Where do I need to be careful about constant factors and where I can choose slower but simpler implementation?
	----------------------------------
	If not AC:
 
	Did you remember to do everything to handle the tricky places you thought about before?
	Is the solution correct?
	Do I understand the problem correctly?
	----------------------------------
	If you have a test on which the solution gives wrong answer:
 
	Is the solution doing what it was supposed to do?
	Is the problem in the code or in the idea?
*/

Compilation message (stderr)

glo.cpp:61:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
   61 | main(){
      | ^~~~
glo.cpp: In function 'void fp(std::string)':
glo.cpp:13:29: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   13 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                      ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
glo.cpp:13:70: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   13 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                                                               ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 5 / 5
• Subtask 2: 10 / 10
• Subtask 3: 13 / 13
• Subtask 4: 10 / 10
• Subtask 5: 20 / 20
• Subtask 6: 17 / 17
• Subtask 7: 25 / 25