Submission #489526

#	Time	Username	Problem	Language	Result	Execution time	Memory
489526		kiomi	Unique Cities (JOI19_ho_t5)	C++17	100 / 100	458 ms	56432 KiB

joi2019_ho_t5

//#pragma GCC target ("avx2")
//#pragma GCC optimization ("Ofast")
//#pragma GCC optimization ("unroll-loops")

//#pragma comment(linker,"/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

//order_of_key(k): Number of items strictly smaller than k .
//find_by_order(k): K-th element in a set (counting from zero).
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define full(x,n) x,x+n+1
#define full(x) x.begin(),x.end()
#define finish return 0

#define putb push_back
#define f first
#define s second

//logx(a^n)=loga(a^n)/logx(a)
//logx(a*b)=logx(a)+logx(b)
//logx(y)=log(y)/log(x)
//logb(n)=loga(n)/loga(b)

#define ordered_set tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>
#define putf push_front
#define gainb pop_back

//(a+b)^n=sum of C(n,i)*a^i*b^(n-i) 0<=i<=n
//(a-b)^n=sum of C(n,i)*a^i*b^(n-i) for even i-for odd i

#define gainf pop_front
#define len(x) (int)x.size()

// 1/b%mod=b^(m-2)%mod
// (a>>x)&1==0
// a^b=(a+b)-2(a&b)

typedef double db;
typedef long long ll;

//sum of squares n*(n+1)*(2n+1)/6
//sum of cubes [n*(n+1)/2]^2
//sum of squares for odds n*(4*n*n-1)/3
//sum of cubes for odds n*n*(2*n*n-1)

const int ary=1e6+5;
const ll mod=1e9+7;
const ll inf=1e18;

using namespace std;
using namespace __gnu_pbds;
int n,m,c[ary],d[ary],mx1[ary],mx2[ary],ans[ary],res,cnt[ary];
vector<int> g[ary],q;
void dfs(int v,int pr=0){
	d[v]=d[pr]+1;
	mx1[v]=mx2[v]=v;
	for(int i=0;i<len(g[v]);i++){
		int to=g[v][i];
		if(to==pr){
			continue;
		}
		dfs(to,v);
		if(d[mx2[v]]<d[mx1[to]]){
			mx2[v]=mx1[to];
		}
		if(d[mx1[v]]<d[mx2[v]]){
			swap(mx1[v],mx2[v]);
		}
	}
}
void del(){
	res-=--cnt[c[q.back()]]==0;
	q.pop_back();
}
void add(int v){
	q.push_back(v);
	res+=cnt[c[v]]++==0;
}
void go(int v,int pr=0){
		for(int i=0;i<len(g[v]);i++){
			int to=g[v][i];
			if(to==pr||mx1[to]!=mx1[v]){
				continue;
			}
			while(len(q)&&d[mx2[v]]-d[v]>=d[v]-d[q.back()]){
				del();
			}
			add(v);
			go(to,v);
		}
		for(int i=0;i<len(g[v]);i++){
			int to=g[v][i];
			if(to==pr||mx1[to]==mx1[v]){
				continue;
			}
			while(len(q)&&d[mx1[v]]-d[v]>=d[v]-d[q.back()]){
				del();
			}
			add(v);
			go(to,v);
		}
	while(len(q)&&d[mx1[v]]-d[v]>=d[v]-d[q.back()]){
		del();
	}
	ans[v]=max(ans[v],res);
}
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	for(int i=1;i<=n;i++){
		cin>>c[i];
	}
	dfs(1);
	int A=mx1[1];
	d[A]=0;
	dfs(A);
	go(A);
	while(len(q)){
		del();
	}
	int B=mx1[A];
	d[B]=0;
	dfs(B);
	go(B);
	for(int i=1;i<=n;i++){
		cout<<ans[i]<<"\n";
	}
}

Compilation message (stderr)

joi2019_ho_t5.cpp:15: warning: "full" redefined
   15 | #define full(x) x.begin(),x.end()
      | 
joi2019_ho_t5.cpp:14: note: this is the location of the previous definition
   14 | #define full(x,n) x,x+n+1
      |

• Subtask 1: 4 / 4
• Subtask 2: 32 / 32
• Subtask 3: 32 / 32
• Subtask 4: 32 / 32