This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "traffic.h"
#include <bits/stdc++.h>
#include <string>
#include <iostream>
#include <cmath>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<int, int> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef vector<ld> vd;
typedef vector<long long> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
#define FOR(i, a, b) for (ll i = a; i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i, a, b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i, a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto &a : x)
#define uid(a, b) uniform_int_distribution<int>(a, b)(rng)
#define len(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define F first
#define nl endl
#define S second
#define lb lower_bound
#define ub upper_bound
#define aint(x) x.begin(), x.end()
#define raint(x) x.rbegin(), x.rend()
#define ins insert
const int MOD = 1000000007;
vvl paths(1e6 + 10);
vvl children(1e6 + 10);
vpi level(1e6+10);
vi parent(1e6+10);
void dfs(int pos,int height,int visited){
level[pos] = {height,pos};
for(auto e:paths[pos]){
if(e == visited)continue;
parent[e] = pos;
dfs(e,height+1,pos);
}
}
int LocateCentre(int N, int pp[], int S[], int D[]) {
ll sum = 0;
ll ans = 9e18;
int anspos = -1;
FOR(i,0,N)sum += pp[i];
paths = vvl(N,vl());
children = vvl(N,vl());
level = vpi(N);
parent = vi(N,-1);
FOR(i,0,N-1){
paths[S[i]].pb(D[i]);
paths[D[i]].pb(S[i]);
}
dfs(0,0,-1);
sort(level.rbegin(),level.rend());
FOR(i,0,N){
ll a = level[i].S;
ll temp3 = -1;
ll temp = sum - pp[a], temp2 = pp[a];
for(auto j:children[a]){
temp3 = max(temp3,j);
temp -= j;
temp2 += j;
}
children[a].pb(temp);
temp3 = max(temp3,temp);
if(ans > temp3)anspos = a;
ans = min(ans,temp3);
if(a == 0)continue;
children[parent[a]].pb(temp2);
}
return anspos;
}
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