Submission #1108999

#	Time	Username	Problem	Language	Result	Execution time	Memory
1108999		doducanh	Factories (JOI14_factories)	C++14	100 / 100	3815 ms	357932 KiB

factories

#include <bits/stdc++.h>
//#include "factories.h"
using namespace std;
#define ll long long
#define fi first
#define se second
const ll MOD = 1e9 + 7;
const ll N = 5e5 + 2;
const ll inf = 1e14 + 2;

vector<ll> minn(N);
vector<int> sz(N), used(N, 0);
vector<vector<pair<int, ll>>> adj(N), ancestors(N);

int n;
void find_sz(int node, int parent) {
    sz[node] = 1;
    for (auto child : adj[node]) {
        if (child.first == parent || used[child.first]) continue;
        find_sz(child.first, node);
        sz[node] += sz[child.first];
    }
}

void update(int node, int parent, int c, ll d) {
    ancestors[node].push_back(make_pair(c, d));
    for (auto child : adj[node]) {
        if (child.first == parent || used[child.first]) continue;
        update(child.first, node, c, d + child.second);
    }
}

void find_cent(int node, int parent, int size) {
    for (auto child : adj[node]) {
        if (child.first == parent || used[child.first]) continue;
        if (sz[child.first] * 2 > size) {
            find_cent(child.first, node, size);
            return;
        }
    }
    update(node, -1, node, 0);
    used[node] = 1;
    find_sz(node, -1);
    for (auto child :adj[node]) {
        if (used[child.first]) continue;
        find_cent(child.first, node, sz[child.first]);
    }
}

void Init(int N, int A[], int B[], int D[]) {
    for (int i = 0; i < N-1; i++) {
        adj[A[i]].push_back(make_pair(B[i], D[i]));
        adj[B[i]].push_back(make_pair(A[i], D[i]));
    }
    find_sz(0, -1);
    find_cent(0, -1, N);
}
long long Query(int S, int X[], int T, int Y[])
{
    for(int i=0;i<T;i++){
        for(pair<int,long long> y:ancestors[Y[i]]){
            minn[y.fi]=inf;
        }
    }
    for(int i=0;i<S;i++){
        for(pair<int,ll> y:ancestors[X[i]]){
            minn[y.fi]=min(minn[y.fi],y.se);
        }
    }
    ll ans=inf;
    for(int i=0;i<T;i++){
        for(pair<int,ll> y:ancestors[Y[i]]){
            ans=min(ans,minn[y.fi]+y.se);
        }
    }
    return ans;
}

• Subtask 1: 15 / 15
• Subtask 2: 18 / 18
• Subtask 3: 67 / 67