# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
924962 | pan | Power Plant (JOI20_power) | C++17 | 0 ms | 0 KiB |
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 <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#include "bits_stdc++.h"
#include <stdio.h>
#include <algorithm>
#include <memory.h>
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
using namespace std;
//using namespace __gnu_pbds;
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
ll n, x, y, answer = 0;
string power;
vector<ll> adj[200005];
ll dp[200005], ans[200005];
// dp -> if 0, 1 can be chosen freely under this subtree
vector<ll> visited;
void dfs(ll from)
{
visited[from] = true;
ll summ = 0, maxx = 0;
for (ll u: adj[from])
{
if (visited[u]) continue;
dfs(u);
summ+=dp[u];
maxx = max(maxx, dp[u]);
}
if (power[from-1]=='1') {ans[from] = max(summ-1, maxx + 1); dp[from] = max(1LL, summ-1);}
// to make below "activated valid, current from must be broken -> summ-1
// else all must be disabled, with only current from left activated
else {ans[from] = dp[from] = summ;}
answer = max(answer, ans[from])
}
int main()
{
input(n);
for (ll i=0; i<n-1; ++i)
{
input(x); input(y);
adj[x].pb(y);
adj[y].pb(x);
}
cin >> power;
visited.assign(n+1, false);
dfs(1);
print(answer, '\n');
return 0;
}