Submission #802813

#	Time	Username	Problem	Language	Result	Execution time	Memory
802813		fatemetmhr	Highway Tolls (IOI18_highway)	C++17	12 / 100	119 ms	16324 KiB

highway

//  ~ Be Name Khoda ~  //

#include "highway.h"
#include <bits/stdc++.h>
//#pragma GCC optimize ("O3")
//#pragma GCC target("avx2")
//#pragma GCC optimize("unroll-loops,Ofast")

using namespace std;

typedef long long ll;

#define pb       push_back
#define mp       make_pair
#define all(x)   x.begin(), x.end()
#define fi       first
#define se       second

const int maxn  =  1e6   + 10;
const int maxn5 =  2e5   + 10;
const int maxnt =  1.2e6 + 10;
const int maxn3 =  1e3   + 10;
const int mod   =  1e9   + 7;
const ll  inf   =  1e18;

vector <int> tok, adj[maxn5], v, u, ver, ed;
ll len;
int m, q[maxn5], par[maxn5];
ll pas[maxn5];
bool mark[maxn5], good[maxn5];

void bfs(int rt){
    par[rt] = -1;
    memset(mark, false, sizeof mark);
    int l = 0, r = 1;
    q[0] = rt;
    mark[rt] = true;
    while(l < r){
        int a = q[l++];
        for(auto id : adj[a]){
            int b = u[id] == a ? v[id] : u[id];
            if(mark[b])
                continue;
            //cout << "here " << a << ' ' << b << ' ' << id << endl;
            good[id] = true;
            mark[b] = true;
            q[r++] = b;
            par[b] = a;
        }
    }
}

void dfs(int v){
    //cout << v << ' ' << par[v] << endl;
    for(auto id : adj[v]) if(good[id]){
        int u = (::u[id]) == v ? (::v[id]) : (::u[id]);
        if(u == par[v])
            continue;
        //cout << "in " << v << ' ' << u << ' ' << id << endl;
        dfs(u);
        ver.pb(u);
        ed.pb(id);
    }
}

ll get(int id){
    if(pas[id] == -1){
        fill(all(tok), 1);
        for(auto u : ed)
            tok[u] = false;
        for(int i = 0; i <= id; i++)
            tok[ed[i]] = true;
        pas[id] = ask(tok);
    }
    return pas[id];
}


void find_pair(int n, std::vector<int> U, std::vector<int> V, int a, int b){
    memset(pas, -1, sizeof pas);
    u = U;
    v = V;
    m = u.size();
    for(int i = 0; i < m; i++){
        adj[u[i]].pb(i);
        adj[v[i]].pb(i);
    }
    int lo = -1, hi = n;
    tok.resize(m);
    fill(all(tok), 0);
    len = ask(tok) / a;
    while(hi - lo > 1){
        int mid = (lo + hi) >> 1;
        fill(all(tok), 0);
        for(int i = 0; i <= mid; i++)
            for(auto id : adj[i])
                tok[id] = 1;
        ll cost = ask(tok);
        if(cost == len * a)
            lo = mid;
        else
            hi = mid;
    }
    assert(hi < n);
    hi = 0; // pay attention, it've changed!!!!!!!!!!!!!!!!!!!
    par[hi] = -1;
    bfs(hi);
    //return;
    dfs(hi);
    /*
    for(int i = 0; i < n - 1; i++)
        cout << ed[i] << ' ' << ver[i] << endl;
    //*/
    //cout << "its " << hi << endl;
    lo = -1; hi = n - 1;
    ll x;
    int lim = 60000;
    if(false && n > lim + 5){
        x = get(lim);
        if(x == len * a){
            lo = lim;
            hi = n - 1;
        }
        else{
            lo = -1;
            hi = lim;
        }
    }
    while(hi - lo > 1){
        int mid = (lo + hi) >> 1;
        if(get(mid) == len * a)
            lo = mid;
        else
            hi = mid;
    }
    int s = ver[hi];
    answer(s, 0);
    return;
    //cout << "here " << lo << ' ' << hi << ' ' << s << endl;
    if(false && n > lim + 5){
        if(x == len * b){
            hi = lim;
            lo = -1;
        }
        else{
            lo = lim;
            hi = n - 1;
        }
    }
    else{
        lo = -1;
        hi = n - 1;
    }
    while(hi - lo > 1){
        int mid = (lo + hi) >> 1;
        //cout << mid << ' ' << get(mid) << ' ' << len << ' ' << a << ' ' << b << endl;
        if(get(mid) == len * b)
            hi = mid;
        else
            lo = mid;
    }
    int v = s;
    bool re = false;
    while(v != -1){
        if(v == ver[hi])
            re = true;
        v = par[v];
    }
    int t;
    if(re)
        t = par[ver[hi]];
    else
        t = ver[hi];
    //cout << re << ' ' << hi << ' ' << lo << ' ' << t << endl;
    answer(s, t);
    return;

}

• Subtask 1: 5 / 5
• Subtask 2: 7 / 7
• Subtask 3: 0 / 6
• Subtask 4: 0 / 33
• Subtask 5: 0 / 18
• Subtask 6: 0 / 31