Submission #135897

#	Time	Username	Problem	Language	Result	Execution time	Memory
135897		shashwatchandra	Torrent (COI16_torrent)	C++17	100 / 100	1323 ms	25904 KiB

torrent

/*input

*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
 
#define double long double
#define f first
#define s second
#define mp make_pair
#define pb push_back
 
#define RE(i,n) for (int i = 1; i <= n; i++)
#define RED(i,n) for (int i = n; i > 0; i--)
#define REPS(i,n) for(int i = 1; (i*i) <= n; i++)
#define REP(i,n) for (int i = 0; i < (int)n; i++)
#define FOR(i,a,b) for (int i = a; i < b; i++)
#define REPD(i,n) for (int i = n-1; i >= 0; i--)
#define FORD(i,a,b) for (int i = a; i >= b; i--)
 
#define all(v) v.begin(),v.end()
#define pii pair<int,int>
#define vi vector<int>
#define vvi vector<vi>
#define print(arr) for (auto it = arr.begin(); it != arr.end(); ++it) cout << *it << " "; cout << endl;
#define debug(x) cout << x << endl;
#define debug2(x,y) cout << x << " " << y << endl;
#define debug3(x,y,z) cout << x << " " << y << " " << z << endl;
 
typedef tree<
int,
null_type,
less<int>,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
 
const int MOD = 1e9+7;
const double PI = 3.14159265358979323846264338;
 
int raise(int a,int n,int m = MOD){
  if(n == 0)return 1;
  if(n == 1)return a;
  int x = 1;
    x *= raise(a,n/2,m);
    x %= m;
    x *= x;
    x %= m;
    if(n%2)x*= a;
    x %= m;
    return x;
}
 
int floor1(int n,int k){
    if(n%k == 0 || n >= 0)return n/k;
    return (n/k)-1;
}
 
int ceil1(int n,int k){
    return floor1(n+k-1,k);
}
 
const int N = 3e5+1;
int n,x,y;
vector<int> adj[N];
int vis = 0;
vector<pii> edges;
int par[N];
int F1,F2;
 
int trynow(int u,int p){
	
	vector<int> childs;
	for(int v:adj[u]){
		if(v == p)continue;
      	if(min(u,v) == min(F1,F2) and max(u,v) == max(F1,F2))continue;
		childs.pb(trynow(v,u));
	}
	sort(all(childs));reverse(all(childs));
	int res = 0;
	REP(i,childs.size()){
		res = max(res,1+i+childs[i]);
	}
	return res;
}

void dfs0(int u,int p){
	par[u] = p;
	for(int v:adj[u]){
		if(v == p)continue;
		dfs0(v,u);
	}
}

bool res(int i){
	// deletes ith edge and tells which dp is greater
	// 1 means x one is greater and 0 means other one
	F1 = edges[i].f;
    F2 = edges[i].s;
	int one = trynow(x,0);
	int two = trynow(y,0);
	F1 = 0;F2 = 0;
	return (one >= two);
}

int giveans(int i){
	F1 = edges[i].f;
    F2 = edges[i].s;
	int one = trynow(x,0);
	int two = trynow(y,0);
	F1 = 0;F2 = 0;
	return max(one,two);
}
 
void solve(){
  	cin >> n >> x >> y;
  	REP(i,n-1){
  		int a,b;
  		cin >> a >> b;
  		adj[a].pb(b);
  		adj[b].pb(a);
  	}
  	dfs0(x,0);
  	int u = y;
  	while(u != x){
  		edges.pb({par[u],u});
  		u = par[u];
  	}
  	reverse(all(edges));
  	
  	// now as we keep deleting the edge which
  	// is further down dp of x will increase
  	// and dp of y will decrease
  	// we want first A such that sign_dp(A) and sign_dp(A+1) are opposit
  	if(res(0) or edges.size() == 1){
  		cout << giveans(0);
  		return;
  	}
  	int lo = 0;
  	int hi = edges.size();
  	hi -= 2;
  	while(lo <= hi){
  		int mid = (lo+hi)/2;
  		if(res(mid) != res(mid+1)){
  			//cout << mid << endl;
  			cout << min(giveans(mid),giveans(mid+1));
  			return;
  		}
  		else if(res(mid))hi = mid-1;
  		else lo = mid+1;
  	}
}
 
signed main(){
  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
  //freopen(".in","r",stdin);freopen(".out","w",stdout);
  int t = 1;
  //cin >> t;
  while(t--){
    solve();
  }
  return 0;
}

• Subtask 1: 31 / 31
• Subtask 2: 69 / 69