×

Hard

# PREREQUISITES:

Number theory, Segment Trees, Heavy Light Decomposition

# PROBLEM:

You're given a tree of N vertices. Each vertex has a number written on it. Process two types of queries :

1. Add d (input) to numbers of all vertices along the unique path between vertices u and v (both input)
2. Find out GCD of all numbers on the vertices along unique path between vertices u and v (both input).

# QUICK EXPLANATION:

Use Heavy Light Decomposition of the tree to reduce problem to only linear paths. Use the fact that gcd(a, b, c, d..) = gcd(a, b-a, c-b, d-c...) to change segment updates to point updates.

# DETAILED EXPLANATION:

It's a difficult problem and as with many other difficult problems as well - the secret of solving this problem is solving a simplification of the problem.

Simplification : Given tree is a linear path
So what can we do if given tree is a linear path? Now our change operations is reduced to to addition of a constant to a contiguous sub-array and our find opertion reduces to finding out gcd of all numbers on a contiguous sub-array. Experienced programmers would've seen immediately that a datastructure like segment tree with technique of lazy propogation could be useful for this problem, however what information is to be stored at each segment, that is not obvious. There are two very simple identities at the core of this problem. Assume (a,b) represents gcd(a,b).

1. (a,b) = (a,b-a)
2. (a,b,c) = (a,(b,c)) = ((a,b),c)

It is still not clear how these help us. Let us first try to see how gcd is changed (or rather how it is not changed) when we increment all the numbers of a sequence.

Say G = (A1,A2,...,An) and G' = (A1 + k, A2 + k, ... An + k)
Now using identities 1 and 2 we can say that : G is same as (A1, A2 - A1, A3 - A2..., An-An-1)
=> G = (A1, (A2-A1, A3-A2...., An- An-1))

So similarly G' = ( A1 + k, A2 + k - (A1 + k) , ...., An + k - (An-1 + k) )
=> G' = ( A1 + k, A2 - A1, A3 - A2, ..... An - An-1 )
=> G' = ( A1 + k, (A2-A1, A3-A2, ....., An - An-1 ))

Now call { A2-A1, A3-A2, ...., An - An-1 } as the difference sequence. So if we compare G and G', we see that much of the information has been retained through the gcd of the difference sequence. In particular it tells us what information we need to store at each segment of segment tree: first number of the sequence and the gcd of the difference sequence. With this information, its easy to find gcd of the segment : its the gcd of first number and gcd of difference sequence.

It's also easy to increment all numbers of the sequence by d, gcd of difference sequence remains unchanged and the first number increments by d. If we want to merge two consecutive smaller segments to make a larger segment, as we'd need to do with segment trees, we also need to store the last number of the segment.

Using this, it is easy to write a segment tree which can do all operations in O(logN) time, making it an O( (N + Q) * logN) solution.

Back to original problem:

We can solve original problem with the help of a technique called Heavy Light Decomposition of a tree. Essential idea is as follows. We divide the vertex set of the tree in O(N) linear chains such that following property holds : On the unique path between any two vertices u and v, there are utmost 2 * log(N) different chains. If we can find such a decomposition, we could build individual segment trees on each of these chains. Now a single query would correspond to quries on each of these O(log N) chains making a single query O( (log N)2) overall.

I'm not going to explain Heavy Light Decomposition here - rather I'd leave you with an exceptionally well written description of this technique. I myself learnt this technique from here.

# SETTER'S SOLUTION:

Can be found here.

# TESTER'S SOLUTION:

Can be found here.

# REFERENCES:

WCIPEG wiki

This question is marked "community wiki".

1243837
accept rate: 0%

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 1 Tester's solution solves only a simplified version of the problem. answered 11 Jul '12, 18:48 6★num314 16●1 accept rate: 0% 1 Thanks for pointing it out. Now the link has been fixed. (12 Jul '12, 10:33)
 0 Want to add some information and to know other users ideas about C++ solution. I realized author's approach on the contest and get TLE, then i got rid of "vector <>"s and finally add static array and used it to allocate memory for segment trees for any path. answered 12 Jul '12, 15:58 4★jughead 1●1 accept rate: 0% Well it isn't necessary to dinamically allocate memory, since the total length is always the same, you could keep the sizes of each segment tree and keep everything in only one array. I did that, and it passed. For QTREE3 I also needed this optimization, guess it is always a good idea to do it this way. (13 Jul '12, 07:12) MarioYC5★
 0 I am getting TLE!! Can anyone HELP me ? my code : https://ideone.com/m7wSJ0 answered 17 Feb '16, 22:19 1 accept rate: 0%
 0 what is the worst case that cause TLE?, i use HLD + segment tree + lazy update answered 16 May '17, 16:47 3★whyzker 1 accept rate: 0%
 0 Setters's solution gives WA. answered 05 Feb '18, 10:05 1 accept rate: 0%
 0 Those getting TLE, don't use long long int. answered 20 Jul '18, 21:42 1 accept rate: 0%
 0 I am getting WA with a pretty standard lazy propagation. My code:  /* */ #include #include #include #include #include #include #define MAXK (18) #define N (1 << 16) #define MAXE (N<<1) using namespace std; typedef int node_type; typedef int size_type; typedef long long value_type; //typedef int value_type; #define L(v) ((v)<<1) #define R(v) (1|L(v)) #define Q (N << 5) #define root (0) #define null (-1) #define BIT(k) (1ULL<<(k)) #define MASK(k) (BIT(k)-1ULL) typedef unsigned long long u64; value_type global_h[Q],global_ans[Q],global_first[Q],global_last[Q],global_g[Q]; inline u64 enc( u64 x, u64 y, u64 z ) { return x | (y< p; public: size_type id; chain() {} inline node_type pos2node( const size_type pos ) const { return global_chain[pos+(id?_left[id-1]:0)]; } inline size_type node2pos( const node_type x ) const { return global_lookup[x]; } inline void append( node_type x ) { global_lookup[x]= llen[id]++, global_chain[_left[id]++]= x; } chain( size_type id ) { this->id= id, llen[id]= 0, _left[id]= id?_left[id-1]:0; } inline size_type len() const { return llen[id]; } }; class ctree { node_type *adj[N]; value_type w[N]; size_type n,K,d[N],tin[N],tout[N],tick,cur[N]; node_type anc[N][MAXK]; void dfs( node_type x ) { tin[x]= ++tick; for ( size_type k, i= 0; i < cur[x]; ++i ) { node_type y= adj[x][i]; for ( anc[y][0]= x, k= 1; anc[y][k-1] != null; anc[y][k]= anc[anc[y][k-1]][k-1], ++k ) ; d[y]= d[x]+1, dfs(y); } tout[x]= ++tick; } node_type up( node_type x, size_type u ) const { for ( size_type k= 0; u; u>>= 1, ++k ) if ( u & 1 ) x= anc[x][k]; return x; } public: inline void reserve( node_type x, size_type card ) { adj[x]= (node_type *)malloc(card*sizeof *adj[x]); } node_type lca( node_type x, node_type y ) const { if ( tin[x] <= tin[y] && tout[y] <= tout[x] ) return x; if ( tin[y] <= tin[x] && tout[x] <= tout[y] ) return y; if ( d[x] > d[y] ) return lca(up(x,d[x]-d[y]),y); if ( d[y] > d[x] ) return lca(y,x); if ( x == y ) return x; for ( size_type k= K-1; k; --k ) { assert( anc[x][k] == anc[y][k] ); if ( anc[x][k-1] != anc[y][k-1] ) x= anc[x][k-1], y= anc[y][k-1]; } return anc[x][0]; } inline size_type size() const { return n; } inline void assign_weight( node_type x, value_type wgt ) { w[x]= wgt; } inline value_type weight( const node_type x ) const { return w[x]; } node_type level_ancestor( const node_type x, const size_type k ) const { return up(x,k); } const pair children( const node_type x ) const { return {adj[x],cur[x]}; } void init( const size_type n ) { this->n= n, tick= -1; for ( K= 0; (1<weight(ch->pos2node(i)); return ; } auto k= (i+j)>>1; build(L(v),i,k), build(R(v),k+1,j); push_up(v,i,j); } value_type _query( size_type v, size_type i, size_type j, size_type qi, size_type qj ) { push_down(v,i,j); if ( qi > j || qj < i ) return 0LL; if ( qi <= i && j <= qj ) //return ans[v]; return GCD(first[v],g[v]); auto k= (i+j)>>1; auto res= GCD(_query(L(v),i,k,qi,qj),_query(R(v),k+1,j,qi,qj)); push_up(v,i,j); return res; } void _update( size_type v, size_type i, size_type j, size_type qi, size_type qj, value_type d ) { push_down(v,i,j); if ( qi > j || qj < i ) return ; if ( qi <= i && j <= qj ) { h[v]+= d; return ; } auto k= (i+j)>>1; _update(L(v),i,k,qi,qj,d), _update(R(v),k+1,j,qi,qj,d); push_up(v,i,j); } public: range_tree( const chain *ch, const ctree *T ) : ch(ch), T(T) { ans= global_ans+(ch->id?4*_left[ch->id-1]:0); h= global_h+(ch->id?4*_left[ch->id-1]:0); first= global_first+(ch->id?4*_left[ch->id-1]:0); last= global_last+(ch->id?4*_left[ch->id-1]:0); g= global_g+(ch->id?4*_left[ch->id-1]:0); build(1,0,ch->len()-1); } value_type query( size_type qi, size_type qj ) { return _query(1,0,ch->len()-1,qi,qj); } void update( size_type qi, size_type qj, value_type d ) { _update(1,0,ch->len()-1,qi,qj,d); } }; class cgraph { node_type to[MAXE]; size_type last[N],next[MAXE],E,n,p[N],card[N]; bool seen[N]; void dfs( node_type x ) { assert( !seen[x] ); seen[x]= true ; for ( size_type i= last[x]; i != null; i= next[i] ) if ( !seen[to[i]] ) p[to[i]]= i, dfs(to[i]); } public: void add_edge( node_type x, node_type y ) { auto i= E++, j= E++; to[i]= y, next[i]= last[x], last[x]= i; to[j]= x, next[j]= last[y], last[y]= j; } void init( size_type n ) { this->n= n, E= 0; for ( node_type x= root; x < n; p[x]= null, seen[x]= false, last[x++]= null ) ; } void preprocess( ctree &t ) { dfs(root); t.init(n); assert( p[root] == null ); for ( node_type x= root; x < n; card[x++]= 0 ) ; for ( node_type x= root+1; x < n; ++x ) ++card[to[p[x]^1]]; for ( node_type x= root; x < n; ++x ) t.reserve(x,card[x]); for ( node_type x= root+1; x < n; ++x ) { assert( p[x] != null ); t.add_arc(to[p[x]^1],x); } } inline size_type size() const { return n; } }; cgraph G; ctree T; class hpd { const ctree *t; chain chains[N]; size_type chlen; size_type which_chain[N],card[N]; node_type best_son[N]; range_tree *rt[N]; size_type dfs( node_type x ) { assert( !card[x] ); best_son[x]= null; const pair children= t->children(x); for ( size_type i= 0; i < children.second; ++i ) { node_type y= children.first[i]; card[x]+= dfs(y); if ( best_son[x] == null || card[y] > card[best_son[x]] ) best_son[x]= y; } return ++card[x]; } void hld( node_type x, node_type from ) { if ( from == null ) chains[chlen++]= chain(++chain_id); which_chain[x]= chain_id, chains[chlen-1].append(x); if ( best_son[x] != null ) hld(best_son[x],x); const pair children= t->children(x); for ( size_type i= 0; i < children.second; ++i ) { node_type y= children.first[i]; if ( y != best_son[x] ) hld(y,null); } } inline const chain &get_chain( const node_type x ) const { return chains[which(x)]; } inline size_type which( node_type x ) const { return which_chain[x]; } void extract_chain( u64 *cpath, size_type &nn, node_type px, node_type x ) { size_type pid= which(px), id; for ( nn= 0; (id= which(x)) != pid; ) { const chain &ch= get_chain(x); cpath[nn++]= enc(id,0,ch.node2pos(x)); x= t->level_ancestor(ch.pos2node(0),1); } const chain &ch= get_chain(x); if ( ch.node2pos(px)+1 <= ch.node2pos(x) ) cpath[nn++]= enc(pid,ch.node2pos(px)+1,ch.node2pos(x)); } u64 x2z[N], y2z[N]; public: void init( const ctree *t ) { this->t= t, chlen= 0, chain_id= -1; for ( auto x= root; x < t->size(); card[x++]= 0 ) ; } void preprocess() { dfs(root), hld(root,null); for ( auto i= 0; i < chlen; ++i ) rt[i]= new range_tree(&chains[i],t); } void update( node_type x, node_type y, value_type val ) { node_type z= t->lca(x,y); size_type x2zlen,y2zlen; extract_chain(x2z,x2zlen,z,x), extract_chain(y2z,y2zlen,z,y); for ( auto i= 0; i < x2zlen; ++i ) { auto id= x2z[i]&MASK(MAXK); auto qi= (x2z[i]>>MAXK)&MASK(MAXK), qj= x2z[i]>>(MAXK+MAXK); //printf("[A] %d --> %d\n",chains[id].pos2node(qi),chains[id].pos2node(qj)); rt[id]->update(qi,qj,val); } for ( auto i= 0; i < y2zlen; ++i ) { auto id= y2z[i]&MASK(MAXK); auto qi= (y2z[i]>>MAXK)&MASK(MAXK), qj= y2z[i]>>(MAXK+MAXK); //printf("[B] %d --> %d\n",chains[id].pos2node(qi),chains[id].pos2node(qj)); rt[id]->update(qi,qj,val); } auto id= which(z); auto qi= get_chain(z).node2pos(z), qj= qi; rt[id]->update(qi,qj,val); //printf("[C] %d --> %d\n",chains[id].pos2node(qi),chains[id].pos2node(qj)); } value_type query( node_type x, node_type y ) { node_type z= t->lca(x,y); size_type x2zlen,y2zlen; extract_chain(x2z,x2zlen,z,x), extract_chain(y2z,y2zlen,z,y); value_type ans= 0LL; for ( auto i= 0; i < x2zlen; ++i ) { auto id= x2z[i]&MASK(MAXK); auto qi= (x2z[i]>>MAXK)&MASK(MAXK), qj= x2z[i]>>(MAXK+MAXK); ans= GCD(ans,rt[id]->query(qi,qj)); } for ( auto i= 0; i < y2zlen; ++i ) { auto id= y2z[i]&MASK(MAXK); auto qi= (y2z[i]>>MAXK)&MASK(MAXK), qj= y2z[i]>>(MAXK+MAXK); ans= GCD(ans,rt[id]->query(qi,qj)); } auto id= which(z); auto qi= get_chain(z).node2pos(z), qj= qi; ans= GCD(ans,rt[id]->query(qi,qj)); return ans; } hpd() {}; }; hpd H; int main() { node_type x,y; size_type i,j,k,n,qr; int ch; value_type wgt; char comm[0x10]; for (;(n= getnum()) >= 0;) { for ( G.init(n), k= 0; k < n-1; x= getnum(), y= getnum(), G.add_edge(x,y), ++k ) ; G.preprocess(T), T.preprocess(); for ( x= root; x < n; wgt= getnum(), T.assign_weight(x++,wgt) ) ; H.init(&T), H.preprocess(); for ( qr= getnum(); qr--; ) { for ( ;(ch= getchar()) != 'F' && ch != 'C'; ) ; x= getnum(), y= getnum(); switch ( ch ) { case 'F': //printf("%lld\n",H.query(x,y)); printf("%lld\n",H.query(x,y)); break ; case 'C': H.update(x,y,getnum()); break ; default: assert( 0 ); } } break ; } return 0; }  I followed the hints here to convert all vectors to static arrays to avoid TLE. I think the problem can also be with the definition of GCD -- what is the gcd of negative numbers, etc? How did you guys handle that aspect? I have seen the author using abs(). answered 07 Sep '18, 02:26 0★se7kazi 1 accept rate: 0%
 0 Can your Link Cut Tree pass this problem? My LCT has TLE. this is my code answered 26 Feb, 19:41 0★isrothy 1 accept rate: 0%
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question asked: 11 Jul '12, 16:33

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last updated: 26 Feb, 19:41