Submission #1108111

#	Time	Username	Problem	Language	Result	Execution time	Memory
1108111		mohammadsam	Dynamic Diameter (CEOI19_diameter)	C++17	100 / 100	235 ms	54120 KiB

diameter

/*
    in the name of god
*/
//#pragma GCC optimize("Ofast,O3,unroll-loops")
//#pragma GCC target("avx,avx2,fma")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,sse4a,avx,avx2,popcnt,tune=native")

#include <bits/stdc++.h>

using namespace std;
#define int long long 
typedef pair<int,int> pii;
typedef pair<long long ,long long> pll;
typedef long long ll ;

#define File          freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define md(x)         (((x%MD)+MD)%MD)
#define all(V)        V.begin(),V.end()
#define setprec(x)    fixed << setprecision(x)
#define Mp(a,b)       make_pair(a,b)
#define len(V)        (int)(V.size())
#define sep           ' '
#define endl          '\n'
#define pb(x)         push_back(x)
#define fi            first
#define sec           second
#define popcount(x)   __builtin_popcount(x)
#define lid           u<<1
#define rid           (lid)|1
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll N = 1e5 + 10,SQ=320,LOG=31;
const ll inf = 2e18 , MD = 1e9 + 7;
int n ,q ,W;
int dis[N];
int st[N<<1],tim=1,L[N],R[N];
vector<pii> g[N];
int par[N];
void tour(int u,int p){
    st[tim] = u;
    L[u] = min(L[u],tim);
    R[u] = max(R[u],tim);
    tim++;
    for(auto h : g[u]){
        if(h.fi!=p){
            tour(h.fi,u);
            st[tim] = u;
            L[u] = min(L[u],tim);
            R[u] = max(R[u],tim);
            tim++;
        }
    }
}
void dfs(int u,int p,int w){
    par[u] = p;
    dis[u] = dis[p] + w;
    for(auto h : g[u]){
        if(h.fi != p){
            dfs(h.fi,u,h.sec);
        }
    }
}
/*
    a,b,c
    0 -> a
    1 -> ab
    2 -> bc
    3 -> abc
*/
struct node{
    int mn,lazy ; 
    int ans[4];
};
node seg[N<<3];
node merge(node a,node b){
    node ans;
    ans.lazy = 0;
    ans.mn = min(a.mn,b.mn);
    ans.ans[0] = max(a.ans[0],b.ans[0]);
    ans.ans[1] = max(a.ans[1],b.ans[1]);
    ans.ans[1] = max(ans.ans[1],a.ans[0] - 2*b.mn);
    ans.ans[2] = max({a.ans[2],b.ans[2],b.ans[0] - 2*a.mn});
    ans.ans[3] = max({a.ans[3],b.ans[3],a.ans[0] + b.ans[2],a.ans[1] + b.ans[0]});
    return ans;
}
void build(int u=1,int l=1,int r=tim){
    if(r-l==1){
        seg[u].mn = dis[st[l]];
        seg[u].ans[0] = dis[st[l]];
        seg[u].ans[1] = seg[u].ans[2] = -dis[st[l]];
        seg[u].ans[3] = 0;
        return ;
    }
    int mid = (l+r)>>1;
    build(lid,l,mid);
    build(rid,mid,r);
    seg[u] = merge(seg[lid],seg[rid]);
}
void shift(int u,int v){
    seg[u].mn += v;
    seg[u].ans[0] += v;
    seg[u].ans[1] -= v;
    seg[u].ans[2] -= v;
    seg[u].lazy += v;
}
void relax(int u){
    shift(lid,seg[u].lazy);
    shift(rid,seg[u].lazy);
    seg[u].lazy = 0;
}
void update(int s,int e,int v,int u=1,int l=1,int r=tim){
    if(e <= s || e <= l || r <= s) return ;
    if(s <= l and r <= e){
        shift(u,v);
        return ;
    }
    relax(u);
    int mid = (l+r)>>1;
    update(s,e,v,lid,l,mid);
    update(s,e,v,rid,mid,r);
    seg[u] = merge(seg[lid],seg[rid]);
}
node get(int s,int e,int u=1,int l=1,int r=tim){
    if(s <= l and r <= e) return seg[u];
    int mid = (l+r)>>1;
    relax(u);
    if(e <= mid) return get(s,e,lid,l,mid);
    if(s >= mid) return get(s,e,rid,mid,r);
    return merge(get(s,e,lid,l,mid),get(s,e,rid,mid,r));
}
pair<pii,int> E[N];
int32_t main() {
    ios_base::sync_with_stdio(false);cout.tie(0);cin.tie(0);
    cin >> n >> q >> W;
    for(int i = 0;i<n-1;i++){
        int u,v,w;
        cin >> u >> v >> w;
        g[u].pb(Mp(v,w));
        g[v].pb(Mp(u,w));
        E[i] = Mp(Mp(u,v),w);
    }
    fill(L,L+n+1,inf);
    dfs(1,1,0);
    tour(1,1);
    build();
    
    int lst = 0;
    for(int i = 0;i<q;i++){
        int d,w;
        cin >> d >> w;
        d = (d+lst)%(n-1);
        w = (w+lst)%W;
        auto [u,v] = E[d].fi;
        if(par[u] != v) swap(u,v);
        update(L[u],R[u]+1,-E[d].sec + w);
        E[d].sec = w;
        lst = seg[1].ans[3];
        cout << lst << endl;
    }
    return 0;
}

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