Submission #1305905

#	Time	Username	Problem	Language	Result	Execution time	Memory
1305905		Zbyszek99	Designated Cities (JOI19_designated_cities)	C++20	7 / 100	355 ms	52540 KiB

designated_cities

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define ld long double
#define ull unsigned long long
#define ff first
#define ss second
#define pii pair<int,int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define rep(i, b) for(int i = 0; i < (b); ++i)
#define rep2(i,a,b) for(int i = a; i <= (b); ++i)
#define rep3(i,a,b,c) for(int i = a; i <= (b); i+=c)
#define count_bits(x) __builtin_popcountll((x))
#define all(x) (x).begin(),(x).end()
#define siz(x) (int)(x).size()
#define forall(it,x) for(auto& it:(x))
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
//mt19937 mt;void random_start(){mt.seed(chrono::time_point_cast<chrono::milliseconds>(chrono::high_resolution_clock::now()).time_since_epoch().count());}
//ll los(ll a, ll b) {return a + (mt() % (b-a+1));}
const int INF = 1e9+50;
const ll INF_L = 1e18+40;
const ll MOD = 1e9+7;

struct edge
{
    int v;
    ll d1,d2;
};

vector<edge> graph[200001];
ll ansT[200001];
ll dist[200001];
pii best;
int pre[200001];
int maxpre[200001];
int preind[200001];
pll up_edge[200001];
bool was[200001];
int n;
int cur_pre;

ll dfs_sum(int v, int pop)
{
    ll sum = 0;
    pre[v] = cur_pre++;
    preind[pre[v]] = v;
    maxpre[v] = pre[v];
    forall(it,graph[v]) if(it.v != pop) 
    {
        up_edge[it.v] = {v,it.d1};
        dist[it.v] = dist[v]+it.d1;
        sum += it.d2+dfs_sum(it.v,v);
        maxpre[v] = maxpre[it.v];
    }
    return sum;
}

void dfs_ans1(int v, int pop, ll ans)
{
    ansT[1] = max(ansT[1],ans);
    forall(it,graph[v]) if(it.v != pop) dfs_ans1(it.v,v,ans-it.d2+it.d1);
}

void dfs_best(int v, int pop, ll d)
{
    best = max(best,{d,v});
    forall(it,graph[v]) if(it.v != pop) dfs_best(it.v,v,d+it.d1);
}

int find_best(int v)
{
    best = {-1,-1};
    dfs_best(v,v,0);
    return best.ss;
}

const int tree_siz = 1024*512-1;
pll tree_max[tree_siz+1];
ll oper[tree_siz+1];

void add_seg(int akt, int p1, int p2, int s1, int s2, ll x)
{
    if(p2 < s1 || p1 > s2) return;
    if(p1 >= s1 && p2 <= s2)
    {
        tree_max[akt].ff += x;
        oper[akt] += x;
        return;
    }
    tree_max[akt*2].ff += oper[akt];
    tree_max[akt*2+1].ff += oper[akt];
    oper[akt*2+1] += oper[akt];
    oper[akt*2] += oper[akt];
    oper[akt] = 0;
    add_seg(akt*2,p1,(p1+p2)/2,s1,s2,x);
    add_seg(akt*2+1,(p1+p2)/2+1,p2,s1,s2,x);
    tree_max[akt] = max(tree_max[akt*2],tree_max[akt*2+1]);
}

void solve(int v)
{
    cur_pre = 0;
    dist[v] = 0;
    ll s = dfs_sum(v,v);
    rep2(i,tree_siz/2+1,tree_siz)
    {
        tree_max[i].ss = i-(tree_siz/2+1);
    }
    for(int i = tree_siz/2; i >= 1; i--) tree_max[i] = max(tree_max[i*2],tree_max[i*2+1]);
    rep(i,tree_siz+1) tree_max[i].ff = 0;
    rep2(i,1,n) add_seg(1,0,tree_siz/2,pre[i],pre[i],dist[i]);
    rep2(i,1,n) was[i] = 0;
    was[v] = 1;
    rep2(j,2,n)
    {
        s += tree_max[1].ff;
        ansT[j] = max(ansT[j],s);
        int v2 = preind[tree_max[1].ss];
        add_seg(1,0,tree_siz/2,pre[v2],pre[v2],-1e17);
        while(!was[v2])
        {
            add_seg(1,0,tree_siz/2,pre[v2],maxpre[v2],-up_edge[v2].ss);
            was[v2] = 1;
            v2 = up_edge[v2].ff;
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //random_start();
    cin >> n;
    ll sum = 0;
    rep(i,n-1)
    {
        int a,b;
        ll d1,d2;
        cin >> a >> b >> d1 >> d2;
        graph[a].pb({b,d1,d2});
        graph[b].pb({a,d2,d1});
        sum += d1+d2;
    }
    dfs_ans1(1,1,dfs_sum(1,1));
    int p1 = find_best(1);
    int p2 = find_best(p1);
    solve(p1);
   // solve(p2);
    int q;
    cin >> q;
    rep(qq,q)
    {
        int e;
        cin >> e;
        cout << sum - ansT[e] << "\n";
    }
}

• Subtask 1: 0 / 6
• Subtask 2: 7 / 7
• Subtask 3: 0 / 9
• Subtask 4: 0 / 17
• Subtask 5: 0 / 17
• Subtask 6: 0 / 44