Submission #1112815

#	Time	Username	Problem	Language	Result	Execution time	Memory
1112815		InvMOD	Harbingers (CEOI09_harbingers)	C++14	50 / 100	175 ms	62264 KiB

harbingers

#include <bits/stdc++.h>

#define FOR(i, a, b) for(int i = (a); i <= (b); i++)

using namespace std;

using ll = long long;

const int N = 1e5+1;


struct LichaoOptMem{
    struct Line{
        int a; ll b;
        Line(int a = 0, ll b = LLONG_MAX): a(a), b(b) {}
        ll f(ll x){return 1ll*a*x + b;}
        ll slope(){return a;}
        bool operator != (const Line& x) const{
            return (a != x.a || b != x.b);
        }
    };
    vector<bool> rekt;

    vector<int> compress; int global_version = 0;

    vector<vector<pair<Line, int>>> st; int trsz;
    LichaoOptMem(int n = 0, vector<int> comp = vector<int>()): trsz(n){
        compress = comp;
        rekt.resize(n+1);
        st.assign(n+1, vector<pair<Line,int>>(1, make_pair(Line(), 0)));
        return;
    }

    void roll_back(vector<pair<Line,int>>& tr){
        while(rekt[tr.back().second]) tr.pop_back();
        return;
    }

    void update(int l, int r, Line curLine){
        if(l > r) return;
        if(l == r){
            if(st[l].back().first.f(compress[l]) > curLine.f(compress[l]))
                return st[l].push_back(make_pair(curLine, global_version)), void();
        }
        else{
            int mid = l+r>>1;
            roll_back(st[mid]);
            Line version = st[mid].back().first;
            if(version.f(compress[mid]) > curLine.f(compress[mid])) swap(curLine, version);
            if(version.slope() < curLine.slope()) update(l, mid-1, curLine);
            if(version.slope() > curLine.slope()) update(mid+1, r, curLine);
            if(version != st[mid].back().first) st[mid].push_back(make_pair(version, global_version));
        }
        return;
    }

    ll query(int l, int r, ll x){
        if(l > r) return LLONG_MAX;
        int mid = l+r>>1;
        roll_back(st[mid]);
        ll cur = st[mid].back().first.f(x);
        if(x == compress[mid]) return cur;
        if(x < compress[mid]) return min(cur, query(l, mid-1, x));
        if(x > compress[mid]) return min(cur, query(mid+1, r, x));
        return -1;
    }

    ll query(ll x){
        return query(0, trsz, x);
    }

    int addLine(int a, ll b){return ++global_version, update(0, trsz, Line(a, b)), global_version;}

    void rmvLine(int version){return rekt[version] = true, void();}
};

ll dp[N];

vector<pair<int,int>> adj[N];

void dfs(int x, int p, ll cur_dist, LichaoOptMem& lichao, vector<int>& speed){
    if(x) dp[x] += lichao.query(speed[x]) + (1ll * cur_dist * speed[x]);

    int rem_version = lichao.addLine(-cur_dist, dp[x]);

    for(pair<int,int> e : adj[x]){
        int v = e.first, w = e.second;
        if(v != p){
            dfs(v, x, cur_dist + 1ll*w, lichao, speed);
        }
    }
    lichao.rmvLine(rem_version);
    return;
}

void solve()
{
    int n; cin >> n;
    vector<int> speed(n);
    FOR(i, 1, n-1){
        int u,v,w; cin >> u >> v >> w;
        u--, v--;
        adj[u].emplace_back(v, w);
        adj[v].emplace_back(u, w);
    }

    vector<int> compress;
    FOR(i, 1, n-1){
        cin >> dp[i] >> speed[i];
        compress.push_back(speed[i]);
    }

    sort(compress.begin(), compress.end());

    LichaoOptMem lichao((int)compress.size()-1, compress);
    dfs(0, -1, 0, lichao, speed);

    FOR(i, 1, n-1) cout << dp[i] <<" \n"[i == n-1];
}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
	
 	solve();
    return 0;
}

Compilation message (stderr)

harbingers.cpp: In member function 'void LichaoOptMem::update(int, int, LichaoOptMem::Line)':
harbingers.cpp:46:24: warning: suggest parentheses around '+' inside '>>' [-Wparentheses]
   46 |             int mid = l+r>>1;
      |                       ~^~
harbingers.cpp: In member function 'll LichaoOptMem::query(int, int, ll)':
harbingers.cpp:59:20: warning: suggest parentheses around '+' inside '>>' [-Wparentheses]
   59 |         int mid = l+r>>1;
      |                   ~^~

• Subtask 1: 50 / 100