Submission #943937

#	Time	Username	Problem	Language	Result	Execution time	Memory
943937		vjudge1	Dynamic Diameter (CEOI19_diameter)	C++17	0 / 100	124 ms	29528 KiB

diameter

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
#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 N = 1e5 + 100, mxc = 105, inf = 1e18, MX = 1e5 + 10;

set <pair <int,int> > g[N];

pair <int, int> t[N * 4];
int tx[N * 4];
int di[N * 4];

void update(int v){
	if(v == 1)return;
    int par = v / 2;
    tx[par] = max({tx[par], t[v].ff + t[v].ss, tx[v]});
    t[par].ss = max(t[par].ss, t[v].ff + di[v]);
    if(t[par].ff < t[par].ss)swap(t[par].ff, t[par].ss);
    update(par);
}
 
int res=0,mx=0,va;

void dfs(int v,int par = -1,int dis = 0){
	if(dis>mx){
		mx=dis;
		va=v;
	}
	for(auto to:g[v]){
		if(to.ss==par){
			continue;
		}
		dfs(to.ss,v,dis+to.ff);
	}
}

main(){
	iostream::sync_with_stdio	(false);	
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n, q, W;
    cin >> n >> q >> W;
	vector <array<int, 3> > vp(n);
	set <pair <int,int> > st;
	for(int i = 0; i < n - 1; i++){
		int a, b, c;
		cin >> a >> b >> c;
		g[a].insert({c, b});
		g[b].insert({c, a});
		di[b] = c;
		st.insert({c, b});
		vp[i] = {a, b, c};
	}
	for(int i = n; i >= 1; i--){
		update(i);
	}
	int last = 0;
	while(q--){
		int d, e;
		cin >> d >> e;
		d = (d + last) % (n - 1);
		e = (e + last) % W;
		auto &[a, b, c] = vp[d];
		g[a].erase({c, b});
		g[b].erase({c, a});
		st.erase({c, b});
		di[b] = e;
		g[a].insert({e, b});
		g[b].insert({e, a});
		vp[d][2] = e;
		if(b / 2 != a)return EXIT_FAILURE;
		update(b);
		cout << max(tx[1], t[1].ff + t[1].ss) << endl;
		last =  max(tx[1], t[1].ff + t[1].ss);
		/*else{
			dfs(1);
			dfs(va);
			cout << mx << endl;
			last = mx;
			mx = -inf;
		}
		*/ 
	}
}	
/*
 * 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)

diameter.cpp:52:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
   52 | main(){
      | ^~~~
diameter.cpp: In function 'void fp(std::string)':
diameter.cpp:15:29: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                      ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
diameter.cpp:15:70: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                                                               ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 0 / 11
• Subtask 2: 0 / 13
• Subtask 3: 0 / 7
• Subtask 4: 0 / 18
• Subtask 5: 0 / 24
• Subtask 6: 0 / 27