/*input
17 1 2
1 3
1 4
4 6
6 7
3 8
3 9
3 10
1 13
13 5
13 11
13 12
13 14
14 15
15 16
15 17
14 2
*/
#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;
set<int> adj[N];
int vis = 0;
vector<pii> edges;
int par[N];
int trynow(int u,int p){
vis--;
if(adj[u].size() < 2 and p)return 0;
vector<int> childs;
for(int v:adj[u]){
if(v == p)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
adj[edges[i].f].erase(edges[i].s);
adj[edges[i].s].erase(edges[i].f);
int one = trynow(x,0);
int two = trynow(y,0);
adj[edges[i].f].insert(edges[i].s);
adj[edges[i].s].insert(edges[i].f);
return (one >= two);
}
int giveans(int i){
adj[edges[i].f].erase(edges[i].s);
adj[edges[i].s].erase(edges[i].f);
int one = trynow(x,0);
int two = trynow(y,0);
adj[edges[i].f].insert(edges[i].s);
adj[edges[i].s].insert(edges[i].f);
return max(one,two);
}
void solve(){
cin >> n >> x >> y;
REP(i,n-1){
int a,b;
cin >> a >> b;
adj[a].insert(b);
adj[b].insert(a);
}
dfs0(x,0);
int u = y;
while(u != x){
edges.pb({par[u],u});
u = par[u];
}
reverse(all(edges));
REP(i,edges.size()){
//cout << edges[i].f << " " << edges[i].s << endl;
}
//cout << res(2) << endl;
// 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;
int its = 0;
while(lo <= hi){
its++;
assert(its < 14);
int mid = (lo+hi+1)/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;
}
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
17 ms |
14584 KB |
Output is correct |
| 2 |
Correct |
16 ms |
14456 KB |
Output is correct |
| 3 |
Correct |
17 ms |
14584 KB |
Output is correct |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Execution timed out |
2033 ms |
46584 KB |
Time limit exceeded |
| 2 |
Halted |
0 ms |
0 KB |
- |
